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.
@@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
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.
actual metric: equation is derived off of s=ut + 1/2 * a * t^2 u is initial velocity and therefore 0 a is 9.81 as it is acceleration due to gravity t is time taken to fall s is displacement (although here it is just the distance fallen)
The equation for gravity is -(g/2)x^2+ vx+h, with g being the gravitational acceleration of whatever body you’re on, v being the initial vertical velocity, and h being the initial height. The gravity of earth is approximately 32 feet per second squared, and the rock took 2.65 seconds to reach ground, and had no initial velocity, meaning the equation is -16(2.65)^2+h=0, and solve for h So -16(2.65)^2=-h The negatives cancel out so 16(2.65)^2=h So time squared * 16 to find initial height
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.
@@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.
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.
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 :)
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)
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.
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.
Don’t you still need to measure the weight of the rock because the heavier the rock the faster it drops which means you will have a less accurate height and the same thing with the lighter one it will drop slower meaning you are still less accurate so my question is how heavy does the rock have to be
@MarkRober You're wrong. You have already calculated the height of the handrail of the bridge, the data on the internet only calculates from where you are standing on the bridge to the water surface, that's why you are wrong.
Yooooo I know that place! That’s an excellent swimming spot (at least it used to be i havent been in a while), called Bridge to Nowhere in Southern California. Super cool, I highly reccomend.
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!
Someone mention that average reaction time is 0,24s for this bias in the measure would be 6.52 meters. Please be aware that this bias is rising with the height of the measured distance. As velocity is rising with time.
I have a question Mark, does this take into consideration of the acceleration of the rock? If so, wouldn't a larger or smaller rock give you a different result?
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 😮
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.
And the speed of sound isn't infinite. But I guess the delay is also negligible
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!!!!
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
This is exactly what I was looking for, thank you!
Why did you multiply 3.28
Explain
@@samelliott8355no problem👍👍
@@All_shape_trackShakya To convert metres to feet
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
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
Unrelated but thank you for using enyas music it’s so underrated
“I’m about to jump if this bridge”
Me: still not crazier than making a heist for squirrels.
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.
actual metric:
equation is derived off of s=ut + 1/2 * a * t^2
u is initial velocity and therefore 0
a is 9.81 as it is acceleration due to gravity
t is time taken to fall
s is displacement (although here it is just the distance fallen)
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
0
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
The equation for gravity is -(g/2)x^2+ vx+h, with g being the gravitational acceleration of whatever body you’re on, v being the initial vertical velocity, and h being the initial height. The gravity of earth is approximately 32 feet per second squared, and the rock took 2.65 seconds to reach ground, and had no initial velocity, meaning the equation is
-16(2.65)^2+h=0, and solve for h
So -16(2.65)^2=-h
The negatives cancel out so
16(2.65)^2=h
So time squared * 16 to find initial height
Bro has the slowest reaction time ever 😭😭💀
How?
Have you seen matpat?
Probably the editing ngl
The video's reversed in the beginning bro💀
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.
Keep making us smarter mark 👍
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 😂
"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?
The very only moment when knowledge at school was useful to you in life:
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.
He is the best engineer of this generation !!
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.
This song was always playing in my preschool
Cómo se llama?
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 i have a question
Does the mass of the rock matter?
Air resistance be like: "Am I a joke to you?"
Lol, air resistance is negligible here.
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
The 7 mile hike up to that point is the best
Fr the brighe to nowhere is wild
Yes but I don’t think you fall the last 10 feet, I think that’s the safety buffer for you
Mark Rober makes Physics understandable for everyone!
We love you Mark!!!🎉
As soon as mark fell he instantly regretted it 😂😂
Thanks a lot! That changed my life 😊
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.
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 😂
it’s all fun and games until you accidentally let go of your phone instead of the rock
I’m terrified of bungee jumping.
So am I?!
Me too
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
Bro ain’t no way my math teacher went there and almost hit the bridge
Yoooo that's my nature home. Crazy to see such a popular YT creator feature it. Caught fire near there recently. 😢
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 😆
112.36ft to be exact.
Love the channel Mark. 👍
That was some ROCKY bungee jumping😂
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
Around 112.36 feet according to the formula he gave us
I was NOT expecting to hear Enya there
dont break your spine mark
An engineer rounding to 16 is the most accurate part of this video
what is song
Oh, you were so smooth here Mark...so cool, so calm.
😄👏🏼🌟🌟🌟🌟🌟
glazer
돌의 항력계수를 구해서 공기저항까지 계산해야 합니다
심지어 1초 늦게 스탑했고, 바닥에서 높이측정을 해야 함을 감안 안 했네요.
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 music at the end😂😂
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 😆
Actually there was a small delay between the rock hitting the water and you pressing the stop button.
Then why am I getting 112.36 feet?
🥶🥶🥶🥶you tell him 🥶🥶🥶
Same
Same
ahhh!!! I found someone with the same answer😶🌫️😶🌫️
Thanks for the great video Mark!
@UTTPEmperorMAPPrideshut up
Thank you for loving Mark Rover so far.😔😔
"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.
@@connorross6921no rocks can accelerate faster if they are heavier
The Enya music🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣
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.
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.
How about the height in metric
He forgot that gravity is stronger and weaker in different parts of the world
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)
Вес камня у всех разный значит и скорость падения у всех разная
Скорость падения у всех предметов одинаковая
Hat doesn’t look very high, mark is pretty big compared to the distance between the ground and the bridge.
Of course mark rober forgets about human error because he doesn’t make any😂😂
Wow i am surprised to see your videos in my mother tongue! How is this possible 🤯
I just found out you were gonna be in Yo Gabba Gabba's new series. Can you tell us anything about it? What you will do, experience, etc.
Don’t you still need to measure the weight of the rock because the heavier the rock the faster it drops which means you will have a less accurate height and the same thing with the lighter one it will drop slower meaning you are still less accurate so my question is how heavy does the rock have to be
Bros capping on the height, 103 ft as opposed to “131 ft”
It’s still not clear..
Because the weight of the rock determines it speed from a particular height.
Dont worry about the fall mark, theres water so u wont take fall damage
"Pull the lever, Kronk!" - Famous Last Words
I JUST LEARNED THIS TODAY!!!!!!
@MarkRober You're wrong. You have already calculated the height of the handrail of the bridge, the data on the internet only calculates from where you are standing on the bridge to the water surface, that's why you are wrong.
Happy Diwali mark
Yooooo I know that place! That’s an excellent swimming spot (at least it used to be i havent been in a while), called Bridge to Nowhere in Southern California. Super cool, I highly reccomend.
Instructions unclear. I dropped my phone, pressed start on myself, morphed into a rock with 10 feet, and then cloned myself 16 times.
And if you can't see the rock hitting the surface, it's height is simply "too high".
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!
Someone mention that average reaction time is 0,24s for this bias in the measure would be 6.52 meters. Please be aware that this bias is rising with the height of the measured distance. As velocity is rising with time.
The song brought my childhood back😭
But all rocks are different sizes so some are heavier and will fall faster
No all rocks will fall at the same speed
Mark forgot to test it 10 times and find how uncertain it is smh
I have a question Mark, does this take into consideration of the acceleration of the rock? If so, wouldn't a larger or smaller rock give you a different result?
But isn't it also related to the weight of the stone? The heavier the stone, the slower it falls.
Air resistance, altitude, weight of the rock, latitude, you need to take all these precautions to know the exact distance
What if instead of dropping the rock mistakenly you dropped your phone😅
Учител:Все понял?
Я:Да...
Телефон летящий в реку:За что..?
Mark Rover, thank you for always uploading videos.마크 로버 항상 영상올려주셔서 감사해요
The song listings me thinking men was dead 💀
Accidentally drops phone and presses rock
But if, for example, the stone is very heavy, it's automatically different
Okay, but what do I do after I let go of my cell phone?
Hi dude i love your content
Even a dead chicken looks epic to this kind of music
I was going to invert gravity with a Minecraft command and walk on walls, walking in the air, counting how many feet there are.
Does the weight of the stone affect the calculation, because I think heavier objects fall faster
seriously
이런 물리 계산식 너무 좋아요!!❤ 그런데 실제로 계산해보니 피트 단위로 나오네요 미터 단위로 환산하기 싫으면 중간에 16 이 아닌 4.9를 곱하면 비슷하게 나옵니다.