@@mhmdhojeij2227in vacuum, the weight doesnt affect the dropping speed, and in normal conditions, heavy things like rock can overcome air resistance, so weight doesnt matter
He forgot to account for the 0.24 second delay (which is the average human reaction time) which would put it at 92 feet deep. Since the drop is so short it’s hard to be accurate; the longer the drop time the more accurate this calculation becomes.
I think the delay between the rock being dropped and the timer starting would be less than .24 seconds because he isn’t reacting to the rock being dropped, he drops it himself
The formula is s=ut+1/2at^2. Since he was holding it still and dropped it with not additional force, u is 0 m/s so the formula becomes s=1/2at^2. t^2 is the time in seconds multiplied by itself. 1/2at^2 is half of the acceleration due to gravity on earth (32 f/s = 16 or 9.8 m/s = about 5). If you multiply those values together you get s which stands for displacement - in this case the height which the rock fell. The mass of the object is irrelevant for this formula although technically it doesn’t account for air resistance, but that is negligible.
@@ridog00 This formula doesn’t take external factors such as air resistance into account. It’s basically an ideal scenario, where you have an initial velocity ‘u’ and you accelerate at the rate ‘a’ for ‘t’ seconds. No additional forces. This isn’t the case in the real world though and different objects may experience different amounts of air resistance, this leading to different outcomes even though the formula doesn’t reflect this.
For the people who don’t understand how he calculated that: s=0.5 • a • t^2 s is distance in feet (in this case) a is acceleration in feet per second, the gravitational constant or 32 feet per second t is time in seconds That leaves us with s=16•t^2
@@jumpjump-oz2pr Once the height is great enough air resistance becomes a factor and you have to account for this by using the same equation S = UT + 1/2 AT^2 for that part of the drop. Only acceleration is now zero and starting velocity is the terminal velocity, in this case of the rock.
Since the initial velocity is zero(he dropped it from the rest, a = 9.82 m/s²), the height will be 34.48 meters, which is very close to the height he found: 112.36 feet (34.51 m).
@@mukulsingh3083the speed isn't 9.82 it is the acceleration Speed isn't constant in motion under gravity as there is constant acceleration therefore each second speed increases until it strikes the ground
@@mukulsingh3083There is no 1 "speed" here because its not constant. If you are thinking "velocity = distance / time", thats not accurate here because that only works for constant velocitys when we are assuming its a constant acceloration, not constant velocity, so a linear velocity in respect to time.
The equation to calculate this is very simple and can be very useful in other ways! X=x0+v0*t+0.5*a*t^2 X is the final position X0 is the initial one V0 is the initial speed T is time A is acceleration In this case we can replace a with 9.8 which is earth's gravity and time with the time it took which was 2.6 seconds, or if you don't have access to the time but you know the height you can switch the equation to look like this: X=0+0+0.5*9.8*t^2 30.48/4.9=t^2 √6.22=t 2.5=t (That's assuming the original height was true) Anyway, thanks for reading this :)
@@diy_wizardI can tell u where the 16 came from. I’m studying to be an engineer. The equation is distance= (initial velocity) x (time) + 1/2 x (acceleration) x (time^2). There is 0 initial velocity so we can get rid of that part of the equation. Now he did the time squared so we know that. The 16 comes from when you convert the acceleration due to gravity which is 9.8 meters per second^2 to feet per second^2. There are 3.3 feet to a meter so if you do 3.3 x 9.8 you get 32.34. Now multiply the 1/2 in and you get 16.17. He just rounded it down to 16.
@@creepergaming3445 The SUVAT equation S = UT + (A(T^2))/2 Of course I would still test the jump with a weight similar to my own mass before I did it just to be sure 😆 EDIT: If the rock reaches terminal velocity then you have to use the same equation for that part of the drop, only this time A is zero and U is non-zero so the part of the equation ((A(T^2))/2) disappears instead, leaving S = UT. Multiplying that terminal velocity by the time spent at that velocity yields the distance in metres.
For those of who are still wondering how he calculated, here is what he did: On Y axis, gravity is the only acceleration, which is g= 9.8 m/s² (in Feet is g= 32 ft/s² ) And the time he measured was t= 2.65 seconds From Kinematics equations, we can get distance by D= 1/2 g t² _Calculating in Feet_ D = 1/2 × 32 × (2.65)² D = 16 × (2.65)² { This is what he did } D = 112 ft Approx. [34 m] He rounded it up to 110 ft because of time delay he might have made at the drop.
Proof - h = ut + 1/2 gt² h = 1/2gt². ( u = 0 { no initial momentum , hence ut = 0 }) h = 1/2 × 9.8 × (2.65)² h = 4.9 × 2.65 × 2.65 h = 12.985×2.65 h = 34.41025m or 112.895ft ( approx)
A small note to people out there: do not try this method for bungee jumping if you don't understand how or why it works. If you mess up with the time measurement, even a little, you will hit the ground. The measly difference between 2.5 seconds and 2.65 seconds is *12 whole feet*. Better just tie a decently heavy rock to a long twine and measure it in some other, more reliable, way To those who wish to know why and how this method works, you need a single easy formula from physics: S(t) = vt + at²/2. It states: the distance (S) an object travels in time (t) equals to the sum of its starting velocity (v) multiplied by time (t) and its acceleration (a) multiplied by time squared (t²), divided by two (/2) In this example, we calculate the distance by specifically dropping the rock, not throwing it, because we need the starting velocity to be 0, otherwise the calculation will be way off. Earlier I said that 0.15 second mistake makes up the difference of 12 feet, now we'd be talking 20+ feet for the same mistake When we measure the time, we need to be quite precise, as that time is later multiplied by itself, which means any mistakes made during the measurement get multiplied as well. Time also must be measured in seconds And finally for the acceleration, if the rock was properly dropped, its acceleration will be equal to the acceleration due to gravity, which equals 9.8 m/s² or 32.2 ft/s², this is where "multiply by 16" bit comes from. We multiply the time squared by acceleration, which is 32.2 in the imperial system, and divide it by two, to get roughly 16
Haha I thought the dumbest comment was the guy who actually thought this worked. Now I read your comment 😂😅 wow the stupidity of people now days. He doesn't care about you or your entertainment. He cares about his wallet.
AP Physics I came in clutch: X = Xi + Vt + 1/2at^2 Simplify a bit since we start at position zero and zero velocity: X = 1/2at^2 We know that acceleration is around 9.8 m/s^2 so we can plug that in: X = 1/2(9.8)t^2 Now, just plug in time for T: X = 1/2(9.8)(2.65)^2 This gets us 34.41 meters or 112.89 feet!
Come on Mark, you’re smart. You dropped it from the top of the fence, which added more seconds to the drop. When you fall your not falling from the height your dropped the rock
“You can check the hight of anything using a rock and your phone, I am about to drop the rock and press start on my phone” *drops phone* sir why are you holding a rock and staring at it?
Mark, it's not the most reliable technique because the reaction time you have is not that of an F1 driver, which means that you have a margin of error. To make it more reliable, I would advise putting a sensor on the surface of the water and making sure that as soon as the sensor is activated, the timer stops. After that, you're the engineer.
Remember to do this by sight. If you wait for the sound of a splash or something, then you're adding distance by the time the sound takes to get to you, which is a lot longer than the time the light takes to get to you.
S= gt²/2, g-ускорение свободного падения, S= (9,8*2,65²)/2= 33,3568, но стоит сделать поправку на сопротивления воздуха, в итоге выходит что-то близкое к 30 метрам
the equation was used in this video is S = 1/2at^2 + v0t + h0, because he was stand on the bridge so his v0 is equal 0 and h0 = 0 too, so infer that he just used the equation 1/2at^2
Multiply by 5 for metric!!
Cool!
Thanks for the information, Mark!
helo
Just a little confused. Why multiply by 16? Is that just to figure out the total in feet?
But thats the 10 feet that you dont go. 😅😅
Imagine accidentally dropping your phone while trying to press start time on the rock
My thoughts exactly 😁
Time the phone 🥶
😭
@UTTPEmperorMAPPride what's the problem?
what if you just didn’t hold the phone over the edge 😮
Instructions unclear, I dropped my phone and pressed start on my rock
😂😂😂😂
Instructions unclear, I dropped the phone, pressed myself, and became the rock
Instructions unclear, I became my phone, ate the rock and am now falling from a skyscraper.
The replies 💀😂
Instructions unclear: i dropped myself, my phone, and the rock has hit the second tower
Metric:
(Time x Time) x 5
Eg. (2.65 x 2.65) x 5 = 35 metres
What about the weight of the rock?
@@mhmdhojeij2227in vacuum, the weight doesnt affect the dropping speed, and in normal conditions, heavy things like rock can overcome air resistance, so weight doesnt matter
😅@@mike_not_lodic
Why 16..why 5 ?
How come u get this numbers or why use this number?
Thx!!!!
He forgot to account for the 0.24 second delay (which is the average human reaction time) which would put it at 92 feet deep. Since the drop is so short it’s hard to be accurate; the longer the drop time the more accurate this calculation becomes.
Also the light traveling was not considered.
I think the delay between the rock being dropped and the timer starting would be less than .24 seconds because he isn’t reacting to the rock being dropped, he drops it himself
@@banjeebeeHe is reacting to the rock hitting the water which means he needs to take away average reaction time which is around 0.2 seconds
You’re a smartass 😂
Cool
The formula is s=ut+1/2at^2.
Since he was holding it still and dropped it with not additional force, u is 0 m/s so the formula becomes s=1/2at^2.
t^2 is the time in seconds multiplied by itself. 1/2at^2 is half of the acceleration due to gravity on earth (32 f/s = 16 or 9.8 m/s = about 5).
If you multiply those values together you get s which stands for displacement - in this case the height which the rock fell.
The mass of the object is irrelevant for this formula although technically it doesn’t account for air resistance, but that is negligible.
Thank you!
Thank you so much for this 🤍
so it wouldn’t change if i dropped a 80lb rock and a 5lb rock? also could the density of the rock be able to change the outcome
@@ridog00 This formula doesn’t take external factors such as air resistance into account. It’s basically an ideal scenario, where you have an initial velocity ‘u’ and you accelerate at the rate ‘a’ for ‘t’ seconds. No additional forces.
This isn’t the case in the real world though and different objects may experience different amounts of air resistance, this leading to different outcomes even though the formula doesn’t reflect this.
“I’m about to jump if this bridge”
Me: still not crazier than making a heist for squirrels.
Mark saying “I’m about to jump off this bridge” is surprisingly not the craziest thing he’s said
ممكن متابعه يا أساطير 😢❤
Ikr... I wasn't very surprised either XDDDD
Bro you’re everywhere
@PikwikCarsonsybau
Mark has INSANE dad lore
Impossible lore
Matpat's gonna have to come back to cover all that
*has*
@@ferret6222 Indeed, because he already is a dad
Não faz sentido pq se eu pegar uma pedra mais pesada
@@ferret6222 fixed it didn't know he is a dad
For the people who don’t understand how he calculated that:
s=0.5 • a • t^2
s is distance in feet (in this case)
a is acceleration in feet per second, the gravitational constant or 32 feet per second
t is time in seconds
That leaves us with s=16•t^2
Also derived (no pun intended) from integrating an acceleration vs time graph twice
You just made it not easy again
No air resistance calculated I start flying now
@@jumpjump-oz2pr Once the height is great enough air resistance becomes a factor and you have to account for this by using the same equation S = UT + 1/2 AT^2 for that part of the drop.
Only acceleration is now zero and starting velocity is the terminal velocity, in this case of the rock.
It's actually S = UT + 1/2 AT^2, because U is zero that part disappears.
Bro has the slowest reaction time ever 😭😭💀
How?
Have you seen matpat?
Probably the editing ngl
The video's reversed in the beginning bro💀
"Here's how to measure any cliffs using a rock and you're phone"
I guess if you want, you can also use a human too 💀
@PikwikCarsonI think but a bot
Humans cannot count seconds so precisely.
They’re saying use a human instead of a rock
uuurm i dont think thats what they ment by use a human 💀☠🤣
@PikwikCarsonHow tf did you get 242 subscribers in one week?
As European physics lover: WHY YOU USE FEETS IN CALCULATIONS
American physics also use 9.81m/s² for earth's gravitational acceleration. I think mark was just making it more approachable for his audience
for metres use 4.9 x time^2
Thank you!
Because we’re American.
I know, it’s unfortunate
Yeah I did the calculations in m and got a bit confused until I realised he was talking in ft
H= 0.5g × t²
H = 4.905 ×2.65²
H=34.445m
H= 112.92ft. (16.08×2.65²)
H= ut+1/2at²
Simple kinematical equation yet so useful
Since the initial velocity is zero(he dropped it from the rest, a = 9.82 m/s²), the height will be 34.48 meters, which is very close to the height he found: 112.36 feet (34.51 m).
Can't we use simple speed X time to calculate distance? Speed being 9.82 and time being 2.65 seconds. It comes out to 26.02m?????
@@mukulsingh3083the speed isn't 9.82 it is the acceleration
Speed isn't constant in motion under gravity as there is constant acceleration therefore each second speed increases until it strikes the ground
Yes but its a rough estimate because acceloration is objectively not constant when putting the other factors into account.
@@mukulsingh3083There is no 1 "speed" here because its not constant. If you are thinking "velocity = distance / time", thats not accurate here because that only works for constant velocitys when we are assuming its a constant acceloration, not constant velocity, so a linear velocity in respect to time.
This song was always playing in my preschool
Cómo se llama?
S = ut + 1/2 at²
S = 0 + 0.5×10×2.65²
S = 5 × 7.02
S = 35.10 meter
S = 35.1 × 3.28 feet
S ≈ 115.128 feet
Mark it's actually 112.9 ft approx.
That's what I'm saying. I got 112.36 tho.
@@turbocateyeS475that’s what I got too
@@turbocateyeS475 Same😂
His math isn't mathing 😂
The equation to calculate this is very simple and can be very useful in other ways!
X=x0+v0*t+0.5*a*t^2
X is the final position
X0 is the initial one
V0 is the initial speed
T is time
A is acceleration
In this case we can replace a with 9.8 which is earth's gravity and time with the time it took which was 2.6 seconds, or if you don't have access to the time but you know the height you can switch the equation to look like this:
X=0+0+0.5*9.8*t^2
30.48/4.9=t^2
√6.22=t
2.5=t
(That's assuming the original height was true)
Anyway, thanks for reading this :)
❤ довольно подробно для простого комментария, спасибо
Mark Rober out here doing sidequests 😂😂😂
As soon as mark fell he instantly regretted it 😂😂
I thought you were gonna take 9.81m/s into account
He is not calculating in metric units ;) since I’m European, I don’t know where that 16 came from. Must be something like 0.5g in feet/square second
@@diy_wizardI can tell u where the 16 came from. I’m studying to be an engineer. The equation is distance= (initial velocity) x (time) + 1/2 x (acceleration) x (time^2). There is 0 initial velocity so we can get rid of that part of the equation. Now he did the time squared so we know that. The 16 comes from when you convert the acceleration due to gravity which is 9.8 meters per second^2 to feet per second^2. There are 3.3 feet to a meter so if you do 3.3 x 9.8 you get 32.34. Now multiply the 1/2 in and you get 16.17. He just rounded it down to 16.
@@creepergaming3445 thanks, I was missing the exact conversion between m and ft ^^ btw I am a physicist 🙌
@@creepergaming3445finally someone in the comments that actually knows what they are talking about
@@creepergaming3445 The SUVAT equation S = UT + (A(T^2))/2
Of course I would still test the jump with a weight similar to my own mass before I did it just to be sure 😆
EDIT: If the rock reaches terminal velocity then you have to use the same equation for that part of the drop, only this time A is zero and U is non-zero so the part of the equation ((A(T^2))/2) disappears instead, leaving S = UT. Multiplying that terminal velocity by the time spent at that velocity yields the distance in metres.
For those of who are still wondering how he calculated, here is what he did:
On Y axis, gravity is the only acceleration, which is g= 9.8 m/s² (in Feet is g= 32 ft/s² )
And the time he measured was t= 2.65 seconds
From Kinematics equations, we can get distance by D= 1/2 g t²
_Calculating in Feet_
D = 1/2 × 32 × (2.65)²
D = 16 × (2.65)² { This is what he did }
D = 112 ft Approx. [34 m]
He rounded it up to 110 ft because of time delay he might have made at the drop.
Proof -
h = ut + 1/2 gt²
h = 1/2gt². ( u = 0 { no initial momentum , hence ut = 0 })
h = 1/2 × 9.8 × (2.65)²
h = 4.9 × 2.65 × 2.65
h = 12.985×2.65
h = 34.41025m or 112.895ft ( approx)
It's 25.97mts
The 7 mile hike up to that point is the best
Fr the brighe to nowhere is wild
For anyone wondering the formula is
h=1/2 gt²
Gravity's acceleration is 9.8 m/s² that converted into feet is 32. So it turns into
32/2 * t²
Vote for a metric version of this short ✋🏻
5 * t^2 in metric as a simplified estimate.
1/2gt²=x
@@alexthurgood92 The Gods of physics are angry on you coz you used g=10 m/s^2 :)
Multiply by 4.9 instead of 16
@@random9415 At GCSE level you're taught that acceleration due to gravity is 10 m/s^2, at A-level you're taught it's 9.81 m/s^2.
Air resistance be like: "Am I a joke to you?"
Lol, air resistance is negligible here.
10 feet is also relatively small, every bit counts. Considering that and his reaction time, it was probably 100ft.
Technically the calculation is wrong. The total time is = rock falling + the sound of the rock hitting bottom. The time is longer than actual
Low distance and hight density material... Air resistance is like negligible
∆X = V_i*t + 0.5*a*t^2
V_i = 0 (holding rock at no velocity)
∆X = 0.5*a*t^2
∆X = 0.5*9.81*t^2 (metric)
∆X ≈ 5*t^2 = 5*t*t
That was some ROCKY bungee jumping😂
He is the best engineer of this generation !!
Dear Mark,
If I don't trust the peeps, I'm not using my phone. Nice lesson, tho.
I was NOT expecting to hear Enya there
Mark Rober makes Physics understandable for everyone!
We love you Mark!!!🎉
Here the python code to mesule hight.
😎😎😎😎
a=float(input("wtite the time :"))
b=a**2
c=b*16
print("height :",c)
The music at the end😂😂
Multiply by five and not 16 if you measure with meters like the rest of the world😂
It's for 4.9
Rest of the world that's cute just remember metric never got to the Moon😂
@@rdel714he said rest of the world not of the moon 😂
@@rdel714 i mean imperial units didnt go either, all the math involved in the Apollo programm was done in metric units.
@@amitisraeli1841 As long as the cord wasn't also measured in metres 😂
Bros capping on the height, 103 ft as opposed to “131 ft”
I’m terrified of bungee jumping.
So am I?!
Me too
dont break your spine mark
An simple physics taught me that it never gonna be that acurate
t²*5 gives distance in metres for those not living in the us :)
or if you want to be more accurate, use t*9.8
@@bruce4476 No lol, the formula for distance in function of acceleration is x=v0*t+at²/2, you can't just drop the /2 and ²
The Enya music🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣
A small note to people out there: do not try this method for bungee jumping if you don't understand how or why it works. If you mess up with the time measurement, even a little, you will hit the ground. The measly difference between 2.5 seconds and 2.65 seconds is *12 whole feet*. Better just tie a decently heavy rock to a long twine and measure it in some other, more reliable, way
To those who wish to know why and how this method works, you need a single easy formula from physics: S(t) = vt + at²/2. It states: the distance (S) an object travels in time (t) equals to the sum of its starting velocity (v) multiplied by time (t) and its acceleration (a) multiplied by time squared (t²), divided by two (/2)
In this example, we calculate the distance by specifically dropping the rock, not throwing it, because we need the starting velocity to be 0, otherwise the calculation will be way off. Earlier I said that 0.15 second mistake makes up the difference of 12 feet, now we'd be talking 20+ feet for the same mistake
When we measure the time, we need to be quite precise, as that time is later multiplied by itself, which means any mistakes made during the measurement get multiplied as well. Time also must be measured in seconds
And finally for the acceleration, if the rock was properly dropped, its acceleration will be equal to the acceleration due to gravity, which equals 9.8 m/s² or 32.2 ft/s², this is where "multiply by 16" bit comes from. We multiply the time squared by acceleration, which is 32.2 in the imperial system, and divide it by two, to get roughly 16
Oh, you were so smooth here Mark...so cool, so calm.
😄👏🏼🌟🌟🌟🌟🌟
glazer
Thats bridge to nowhere if anyone is curious
I was looking in comments to confirm this. Only been there once, but this sure looks like it. Anyone curious, it’s in California above Azusa
I’ve been there too, swam in the river just upstream from the bridge
I was looking in the comments to see if anyone else noticed as well. Just bungee jumped off the bridge a few months ago for my 18th
@@John_Doe742 I can just imagine a pile of smashed phones at the bottom now 😆
it’s all fun and games until you accidentally let go of your phone instead of the rock
and now i need to convert that 110 feet to meters
Divide by 3 and a lil more
@@TheEfX Just account for that when you select the length of the bungee cord 😆
Yoooo that's my nature home. Crazy to see such a popular YT creator feature it. Caught fire near there recently. 😢
correct answer was 34,41025
if u need i can write my calculations
Then why am I getting 112.36 feet?
🥶🥶🥶🥶you tell him 🥶🥶🥶
Same
Same
ahhh!!! I found someone with the same answer😶🌫️😶🌫️
"What weight does the rock have to be ?????"
Shouldnt matter every rock should accelerate the same unless you have widely different aerodynamics.
So it will only vary by a few feet each time?
@@u_s_e_rname probably not even that.
12^2 or 45 (for metric) is three seconds deep from Phineas and Ferb maze episode
shout out to mark doing all this just for our enjoyment 🤨‼️
It’s also, you know, his job
Haha I thought the dumbest comment was the guy who actually thought this worked. Now I read your comment 😂😅 wow the stupidity of people now days. He doesn't care about you or your entertainment. He cares about his wallet.
Thanks for the great video Mark!
@UTTPEmperorMAPPrideshut up
Plot twist, rope snaps and you break your skull on the rock
Of course mark rober forgets about human error because he doesn’t make any😂😂
How about the height in metric
Учител:Все понял?
Я:Да...
Телефон летящий в реку:За что..?
What if your on Mars?
Replace the 16 with 4.42
(I made that up but I bet it’s something like that)
@@tee4222 The average gravitational acceleration on Mars is 3.72076 m/s2 (about 38% of the gravity of Earth)
The reason we can ignore any conditions in the formulas of physics is thanks to the efforts of our predecessors.
And if you can't see the rock hitting the surface, it's height is simply "too high".
marks own genius made him freak out a bit
Air resistance, altitude, weight of the rock, latitude, you need to take all these precautions to know the exact distance
AP Physics I came in clutch:
X = Xi + Vt + 1/2at^2
Simplify a bit since we start at position zero and zero velocity:
X = 1/2at^2
We know that acceleration is around 9.8 m/s^2 so we can plug that in:
X = 1/2(9.8)t^2
Now, just plug in time for T:
X = 1/2(9.8)(2.65)^2
This gets us 34.41 meters or 112.89 feet!
Even a dead chicken looks epic to this kind of music
Mark nearly dies pt 1
Mark robers last words: ohh gosh nooooo
it is based on the formula s = ut + 1/2 x at^2, initial speed u is 0 and a = g (roughly 10m/s^2)
Hat doesn’t look very high, mark is pretty big compared to the distance between the ground and the bridge.
Come on Mark, you’re smart. You dropped it from the top of the fence, which added more seconds to the drop. When you fall your not falling from the height your dropped the rock
I know exactly where he is in this video 😂
Fun fact: This is dubbed in multiple languages
THE MUSIC GOT ME😂😂😂😂😂
Buddy yepadi ye ipadi tamil pesura .....😮😮😮😮😮😮😮😮😮நீங்கள் நன்றாக தமிழ் பேசுகிறீர்கள் என்பது எனக்கு ஆச்சரியமாக இருக்கிறது...
Tamil lam pesala dubbed pannirukanga audio track
@@MohamedAshrif-dy9rr yen manathai odaithu vitanga... 😭😭
The thing that scares me the most about bungee jumping is if there are rocks under me i am scared i will hit a rock hard
“You can check the hight of anything using a rock and your phone, I am about to drop the rock and press start on my phone” *drops phone* sir why are you holding a rock and staring at it?
What if instead of dropping the rock mistakenly you dropped your phone😅
It’s still not clear..
Because the weight of the rock determines it speed from a particular height.
Mark, it's not the most reliable technique because the reaction time you have is not that of an F1 driver, which means that you have a margin of error. To make it more reliable, I would advise putting a sensor on the surface of the water and making sure that as soon as the sensor is activated, the timer stops. After that, you're the engineer.
You’re only dropping 100 feet, but that 10 feet is where you’re hovering over
Remember to do this by sight. If you wait for the sound of a splash or something, then you're adding distance by the time the sound takes to get to you, which is a lot longer than the time the light takes to get to you.
S= gt²/2, g-ускорение свободного падения, S= (9,8*2,65²)/2= 33,3568, но стоит сделать поправку на сопротивления воздуха, в итоге выходит что-то близкое к 30 метрам
It’s less the height at that point and more the jagged rocks you’re barreling towards
Its that extra 10 ft that stops you from going splat when you do the bungee jump😂
Greater the mass of the rock, less time it will take to hit the ground. So, is this method accurate. Please explain
so fun games until you drop your phone instead of the rock
Air resistance and time taken by sound to travel, its again 100 feet
Dude I’m throwing a rock off Mount Everest
Instructions unclear, I can’t see the rock no more…
The cameraman: 🎥🦆
This is how Felix Baumgartner also measured the distance?
That has to be the shortest bungie jump ive ever seen.
New Infinite Craft recipe: Rock + Phone = Physics
He forgot that gravity is stronger and weaker in different parts of the world
the equation was used in this video is S = 1/2at^2 + v0t + h0, because he was stand on the bridge so his v0 is equal 0 and h0 = 0 too, so infer that he just used the equation 1/2at^2
Actually there was a small delay between the rock hitting the water and you pressing the stop button.
They probably said 100 feet because thats how far you fall before getting caught by the bungee
Pick the most round rock you will see for extra precision
Wouldn’t it change based off the size of the rock? The rock would fall faster if it was heavier.
Its the last ten feet that might keep me alive