6:15 In this Case Let initial velocity (u) is 20km/hr=5.55m/s Angle is 30° Gravity(g) is 9.8m/s^2 Range that is horizontal distance covered is, R=u^2xsin2(30°)/g =5.55x5.55x0.866/9.8 =2.72 Matlab thele par se gaadi sirf 2.72 meter aage jake giregi. Height attained by scooty is, H=u^2xsin^2(30°)/2g =5.55x5.55x0.25/19.6 =0.4 meter Matlab scooty Gaadi ko jake thokni chahiye. Time Required is, T=2usin(30°)/g =2x5.55x0.5/9.8 =0.6 seconds Matlab ki scooty ko 0.6 seconds me niche girna chahiye par yaha to scooty is flying for 9 seconds💀💀 (1 like to banata hai) Edit: Nishchay Bhai please ♥️
6:41 Given: - Mass of the object (m) = 350 kg - Initial speed (u) = 20 km/h - Angle of projection (θ) = 30° First, we need to convert the initial speed from km/h to m/s: 1 km/h = 1000 m / 3600 s = 5/18 m/s So, initial speed (u) = 20 km/h * 5/18 = 100/18 m/s ≈ 5.56 m/s Now, we can calculate the horizontal and vertical components of the object's velocity: Horizontal component (ux) = u * cos(θ) Vertical component (uy) = u * sin(θ) ux = 5.56 m/s * cos(30°) ≈ 4.81 m/s uy = 5.56 m/s * sin(30°) ≈ 2.78 m/s Next, we can calculate the time of flight (t) using the vertical motion: The time to reach the highest point (time to peak) is given by: t_peak = uy / g, where g = 9.81 m/s² (acceleration due to gravity) t_peak = 2.78 m/s / 9.81 m/s² ≈ 0.283 s Since the motion is symmetrical, the total time of flight is twice the time to reach the peak: t = 2 * 0.283 s = 0.566 s Finally, we can find the horizontal distance (range) the object will travel: Range = ux * t Range = 4.81 m/s * 0.566 s ≈ 2.72 meters Therefore, the object will land approximately 2.72 meters away from the point where it was launched. Answer: The object will land approximately 2.72 meters away from the launch point.
6:37 Velocity = 20 km/hr = 20×1000/3600 = 50/9 m/s = 5.55 m/s Angle of elevation with the ground= 30° Range= (5.55)²sin(2×30°)/(9.8) ≈ 2.72201 metres (Its the maximum range, g is taken constant, and the mass is neglected. If mass is also taken into account then the range would be less than this)
6:33 Initial velocity 20km/h =20x5/18 = 5.5m/s Slope 30° from horizontal axis , varticall velocity hogi usinthetha= 2.75m/s and horizontal velocity hogi ucosthetha =4.8 m/s Ab hamare pass x and y component aa chuke h to easy rahega time of flight niklna. Let's find time of flight= 2u/g = 0.55 s Yaha pr maine Y component ki velocity li h maine cauz (TOF) vertical velocity pr depend krta h na ki horizontal pr. Range nikalte h aab, R=UxT= 4.8x0.55 =2.64m pr jakr land krega scooter Isme horizontal velocity li h cuz range horizontal velocity pr depend krta h.
Oo Bhai gazab sabke answer dekhe Maine sale sab log sidha formula dal ke answer bata rhe the lekin tumne to step wise pura solve kia h and explanation bhi accha kia h jaise vertical velocity hi kyu li time of flight mai 👍
6:38 34.64 m which is not about even above 3 to 4 vehicles . These serial are actually illogical But when u roast them satisfaction milta hai aur hasi bhi band nhi hoti😂😂😂
7:45 Nischay the way u roast 🤣❤️ my goodness u literally make me laugh so much 🤣❤️ Even when we r not in the mood to laugh or not well all we need to do is watch ur vdo ❤️ Thank u so much for making me smile ❤️ U r the best ❤️ Love you Nischay ❤️
6:41 To calculate the distance (range) an object travels when projected at an angle, you can use the formula for the range of a projectile. First, convert the initial speed from km/h to m/s: 1. **Convert initial speed to m/s:** \[ u = 20 \text{ km/h} = \frac{20 \times 1000}{3600} = \frac{20000}{3600} \approx 5.56 \text{ m/s} \] 2. **Apply the formula for the range of a projectile:** \[ R = \frac{u^2 \sin(2\theta)}{g} \] where \( \theta \) is the angle of projection, \( u \) is the initial speed, and \( g \) is the acceleration due to gravity (approximately 9.81 m/s²). 3. **Calculate the range:** - Convert the angle from degrees to radians: \( \theta = 30^\circ = \frac{\pi}{6} \text{ radians} \) - Compute \( \sin(2\theta) \): \[ \sin(2 \times 30^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866 \] - Plug the values into the range formula: \[ R = \frac{(5.56)^2 \times 0.866}{9.81} \] \[ R = \frac{30.8 \times 0.866}{9.81} \approx \frac{26.7}{9.81} \approx 2.72 \text{ meters} \] So, the distance traveled by the object is approximately **2.72 meters**.
Nischay seriously m apko dekh k bhut khush hoti hu life m kbhi apse Milne ka moka mila na m us din bhut khush hongi ap bhut acche ho me kitna hi presan hu pr apko dekh k sach m khush ho jati hu
6:15 In this Case
Let initial velocity (u) is 20km/hr=5.55m/s
Angle is 30°
Gravity(g) is 9.8m/s^2
Range that is horizontal distance covered is,
R=u^2xsin2(30°)/g
=5.55x5.55x0.866/9.8
=2.72
Matlab thele par se gaadi sirf 2.72 meter aage jake giregi.
Height attained by scooty is,
H=u^2xsin^2(30°)/2g
=5.55x5.55x0.25/19.6
=0.4 meter
Matlab scooty Gaadi ko jake thokni chahiye.
Time Required is,
T=2usin(30°)/g
=2x5.55x0.5/9.8
=0.6 seconds
Matlab ki scooty ko 0.6 seconds me niche girna chahiye par yaha to scooty is flying for 9 seconds💀💀
(1 like to banata hai)
Edit: Nishchay Bhai please ♥️
Ooo BC mujhe tution dede vro
Jee aspirant spotted👍👍
@@Swara_47 I am preparing NEET😅😅
@@Harshvardhan_Patil1008 Ohh! Sorry but atleast I was right that you are preparing for a competitive exam ! 😃
💀🫥🙏ram ram bhaiya 🫥🙏
6:41
Given:
- Mass of the object (m) = 350 kg
- Initial speed (u) = 20 km/h
- Angle of projection (θ) = 30°
First, we need to convert the initial speed from km/h to m/s:
1 km/h = 1000 m / 3600 s = 5/18 m/s
So, initial speed (u) = 20 km/h * 5/18 = 100/18 m/s ≈ 5.56 m/s
Now, we can calculate the horizontal and vertical components of the object's velocity:
Horizontal component (ux) = u * cos(θ)
Vertical component (uy) = u * sin(θ)
ux = 5.56 m/s * cos(30°) ≈ 4.81 m/s
uy = 5.56 m/s * sin(30°) ≈ 2.78 m/s
Next, we can calculate the time of flight (t) using the vertical motion:
The time to reach the highest point (time to peak) is given by:
t_peak = uy / g, where g = 9.81 m/s² (acceleration due to gravity)
t_peak = 2.78 m/s / 9.81 m/s² ≈ 0.283 s
Since the motion is symmetrical, the total time of flight is twice the time to reach the peak:
t = 2 * 0.283 s = 0.566 s
Finally, we can find the horizontal distance (range) the object will travel:
Range = ux * t
Range = 4.81 m/s * 0.566 s ≈ 2.72 meters
Therefore, the object will land approximately 2.72 meters away from the point where it was launched.
Answer: The object will land approximately 2.72 meters away from the launch point.
💀💀💀
Broo forget everything about the outer world 💀💀💀
Kaun se grah se ho bhai?🤔
U r great!! 😂
@@cskyoo6495 YOOOO 2 MINS AGO
6:37 Velocity = 20 km/hr = 20×1000/3600 = 50/9 m/s = 5.55 m/s
Angle of elevation with the ground= 30°
Range= (5.55)²sin(2×30°)/(9.8)
≈ 2.72201 metres
(Its the maximum range, g is taken constant, and the mass is neglected. If mass is also taken into account then the range would be less than this)
Sch btana engineering kr rhi ho na😅
Simaraa😅😅
Lemda = 9.8 always remember it by oure science teacher 😊 hehehh by the way wow but in 50/9m/s exact mean I didn't understand!
Hey bro , can u say how much speed and elevation with the ground we need to cover the distance showed in that serial, plss
Sidha answer batana tha
Pura chapter hi pada diya 😂😂
12:19 ,literally 😂i am watching this video while eating bhel puri 😄
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3:42
what a kidney touching acting 😂😂😂
Kuch to touch hua 😂😂
@@_tanya11_😂😂😂
2:16 kawrrrrrrrrrrr 🦴
3:48 gaming insaan set up❤
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6:33
Initial velocity 20km/h =20x5/18 = 5.5m/s
Slope 30° from horizontal axis , varticall velocity hogi usinthetha= 2.75m/s and horizontal velocity hogi ucosthetha =4.8 m/s
Ab hamare pass x and y component aa chuke h to easy rahega time of flight niklna.
Let's find time of flight= 2u/g = 0.55 s
Yaha pr maine Y component ki velocity li h maine cauz (TOF) vertical velocity pr depend krta h na ki horizontal pr.
Range nikalte h aab,
R=UxT= 4.8x0.55 =2.64m pr jakr land krega scooter
Isme horizontal velocity li h cuz range horizontal velocity pr depend krta h.
Oo Bhai gazab sabke answer dekhe Maine sale sab log sidha formula dal ke answer bata rhe the lekin tumne to step wise pura solve kia h and explanation bhi accha kia h jaise vertical velocity hi kyu li time of flight mai 👍
🏅🏆 Le bhai rakh le
you've absolutely smashed it mate.
🤯
Mind blowing sir g 🤓 konsi class mai ho aap 😭
12:26 meri bhi insta feed pura is cheez se ki bharti hainn😂😂
To ye dekh ke feed kyu bhar rhe ho😂😂
10:16 was chikna triggu's supremacy!!❤️❤️
12:05 ese logo ko dekh kar hi india ko unhygienic kahate ha
7:40 whooooo whooooo whoooo😂😂😂❤
5:31 kya pose me statue huye he😂😂
Bhai those marzil and vanjaj ke scenes gave me nostalgic memories back ❤.....
Becuz I'm re-watching this video after completing gaming insaan ❤
9:10 TV serial super cheap 🙄🤒
13:10 Aapne waqt badal diya jasbat badal diya zindagi badal di 😂😂😂😂😂😂 unexpected serials seens ye scam ko India se bhar mat jana do😂😂😂😂😂😂😂😂😂😂😂😂😂
14:21
Didn't realise how these 14 min passed, literally amazing video enjoyed it so much.💖💗
4:01 shaddi bhi ho gya bacha bhi ho gya 😂😂😂😂
10:20 hero toh ho hi aap 😊😊
I
Yeh me kehne hi wali thi fir yeh comment dikh gayi😅
Yes but no my brother
12:11 nischay yeah dekhne ke baad bhelpuri khana mat chod dena 😂😂
12:17 Triggguuuuu....Bhelpuri k sath anyaay ho raha heeee😢😢😢😢😢😢😢😢😢😢...Kuchh to action lena pdegaaa😢😢😂😂
Nonu ne reply de diya kash muje bhi de deta 😢
Iss reel ka link dedo koyi mujhe dkhna hai plzzz
Ye wastage of food hai
Mane aj pahli dfa video dakhi trigger Insan ki mza agya qsm sy what a editing 😂mane aj sy pahly Q na dakhi😂🤌🏻 subscribe to bnta ha ab😂
7:02
😂😂 ro rahe he sab
2:45 ap janata ko murk samjhna band kare ...hagu😂😂❤
Tum kon ho 😮
Etna achha video kiyu bnate ho bhai jishe dhekh ke meri hashi hi nhi rukti hai...😂😅🤣🥰
8:46 Kya hi energy dali he yrr video me❤❤.....Serial ke scene se jyada to aapko dekh k maja aa raha he😅😂...BTW missing your streams😢😢😢
6:55 Omg😂😂😂❤
Funniest part and Funniest video ever😂😂😂❤❤
Bhaiyaa aaapki sb videos ko dekhkr bina hsse rha hi nhi jata😅😅😂😊
13:28 papita ka pate ko lagakar face transplant😂 That make me so laugh 😂😂😂
BRO IN SHOCK 😶😶IN 1SEAND 4:09
Vanshaj 😂😂
0:24 triggered ki hindi ❤😮
2:27 Kya karu me MBBS krna chhod du kyaa😢😢😢😢..Aise hi theek ho jayenge sab🫠🫠🫠🫠🫠
Chod Munna Bhai mbbs wala ban ja😊
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Nischay bhiyaa always hero metrial ♥️🔥🔥🔥🔥🔥🔥😎
4:09 kya scene tha 😂❤ you are the best
7:21 what the heck is that 😶kuch bhi matlab kuch bhi
10:05 i could practically smell the edits
Yeah, I know how shit smells.
Just like trash
10:15 wow omha❤❤❤❤
11:46 That make sense that you are A whole Green forest because you know that who is red forest😁😁😁😁😁❤❤
11:00 Caption gaya tel mein 😶🥲
4:32 Statue 😡🤣
3:56 😂😂😂 41:21 statue 😂😂 12:25 hachiwali seen bohat cute thi yaaaar🤓💞💞appki hachi bohat cute he 12:45 😂😂❤
9:06 sasural simar ka❌
Makkhi✔️
Makhija returns
4:20 Statue 😂😂
Love ur videos😂❤❤ . This 14 min of video made my day❤
That's great
Life hack 069 💀 9:49
Why 69😂
@@joysreeghosh7300 vhii 💀🤣
I am your 100th liker😂
@@rudranarayanmondal1326 are bhai shi hai 😂😂👍🏻
11:49 Nischay be like: pehle khud ko bachalu warna vivaad ho jaega 😂😂
12:50 the mood swings bruhhhh 😂😂😂
POSE THORA CASUAL HE🌚🤧👊🏻 5:40
Tum kon ho
69 likes ☠️💀
@@painfulsadbrokensong3452Chup gawar
@@painfulsadbrokensong3452 insan human😂
sahdeep2005😂50
2:50 Nischay's narration is epic 😂🔥
3:16 avinash mishra jo abhi bigg boss 18 me hai 😂
11:38 phli bat to tujhe uspe bhi gussa ni ana chahiye ayee hayeee gentlemen❤😁🥰🙈😘😘😘😘
12:45 his laugh 🤣🤣🤣🤣🤣🤣🤣🤣
😂😂😂😂
2:30 doctor's be like hum job chorde ?? 🤣
9:40 triggu said daru piyegi ❤😂
6:38 34.64 m which is not about even above 3 to 4 vehicles . These serial are actually illogical
But when u roast them satisfaction milta hai aur hasi bhi band nhi hoti😂😂😂
9:19 logic 😂😂 rip
13:45 -- mujhe toh unse jada apne pe hasi a rahi hai
Pehle ye serial mai seriously dekhti thi😂
Pehle wale serial k naam kya tha?
Sath nibhana sathiya@@RG-lo8wr
@@RG-lo8wr I wanna know the same
13:55 the legend mummy will watching
12:39 aur kha bhelpuri bnd krdo bhai ab ye khana 😂😂
Jhalmuri khao
10:29 use kyu mara 😂😂
7:50 Yaar aise TV serial wale Science ko dictionary mein include hi nahi karte 🤯🤣🔥.
0:57 Shuru ho gaya roast 😂🔥 with physics
Physics m b ek kar rakhi saloo me
1:16
itna achaa toh meri physics ki teacher bhi nahi padhati😂😂
12:55 haas rahe ho ya ro rahe ho 😅
7:45 Nischay the way u roast 🤣❤️ my goodness u literally make me laugh so much 🤣❤️
Even when we r not in the mood to laugh or not well all we need to do is watch ur vdo ❤️
Thank u so much for making me smile ❤️
U r the best ❤️
Love you Nischay ❤️
7:07 Baburao inside Shetty Saab 😂😂
The fact that nobody is talking abt the different ppl and the room😂
* I came here after watching the epic and legendary show *
Theta 30⁰
Velocity 20 km/hours = 5.5 m/s
Range = u² sin2∅/g
Range = 1.5125 √3
=2.619 meter
Kya bhai
Ae baigan 👀🤣🤣
Bhai isko math bolte h na😮
@@LUCKY_GAMERRR😂😂😂
Bhai Bhai
5:03 main aise hi statue hokar phone pakad ke video dekh rahi thi but fir aapki video dekh rahi hu toh hasi toh aa hi jaata hai 😂😂😂
3:56 was epic finally nishchay ki shadi to hui chahe noni banake hui😂😂
Tum kon ho 😮
Happy birthday to triggered insaan bigest fan from Kolkata pls reply me ❤❤
9:33 vo aapaki kaju katli he😂😂
Tum kon ho 😮
6:41
To calculate the distance (range) an object travels when projected at an angle, you can use the formula for the range of a projectile. First, convert the initial speed from km/h to m/s:
1. **Convert initial speed to m/s:**
\[
u = 20 \text{ km/h} = \frac{20 \times 1000}{3600} = \frac{20000}{3600} \approx 5.56 \text{ m/s}
\]
2. **Apply the formula for the range of a projectile:**
\[
R = \frac{u^2 \sin(2\theta)}{g}
\]
where \( \theta \) is the angle of projection, \( u \) is the initial speed, and \( g \) is the acceleration due to gravity (approximately 9.81 m/s²).
3. **Calculate the range:**
- Convert the angle from degrees to radians: \( \theta = 30^\circ = \frac{\pi}{6} \text{ radians} \)
- Compute \( \sin(2\theta) \):
\[
\sin(2 \times 30^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2} \approx 0.866
\]
- Plug the values into the range formula:
\[
R = \frac{(5.56)^2 \times 0.866}{9.81}
\]
\[
R = \frac{30.8 \times 0.866}{9.81} \approx \frac{26.7}{9.81} \approx 2.72 \text{ meters}
\]
So, the distance traveled by the object is approximately **2.72 meters**.
Well done, ChatGPT👍👍
Sidhe pata chal raha hai ki ChatGPT use kara hai. Bhai ek bar check to kar leta🤣🤣
Motion in plane 😅😅 abi whi padh tha tha
Itna Gyan ka show of nahi karte
Chat gpt se liya haii
4:02 Ohhhh vanshaj bhaiyaaa..acting😂😂😂😂...Aap sach me papa ban gye unke😂😅
Papa nhi maama
Maa Kasam jisne like nhibkiya
Kash aaj mere 8 Subscribers hogayen
Hello
@@fahimbhai0068hhh
0:34 best 😂😂
4:05 wah bhaiya kya video hai maja aa gaya 😂😂aise hin saste raho aur hame hasate raho...
4:56 over to karlo
3:51 this scene 😂
❤❤❤❤❤❤😂😂😂😂😊😊😊😅😅😅😅😅😅😅❤❤❤❤❤❤❤❤ hai na 😹 hai 😂
Nischay seriously m apko dekh k bhut khush hoti hu life m kbhi apse Milne ka moka mila na m us din bhut khush hongi ap bhut acche ho me kitna hi presan hu pr apko dekh k sach m khush ho jati hu
3:10 itna inko diya mne😂😂😂
14:10 Most awaited moment 😂😂 as per title: EARTH SE BADI ROTI 😱😅.
EARTH se badi roti❌
MILKY WAY se badi roti☑
Kyunki vah robot thi.... Robot kuch bhi kar sakti h😂😊
0:25 bhaii itni shudh Hindi 😂
Yes
10:14 🥺❤️ Nischay...🫶🏻
1:37 😂😂😂
9:39 When Nischay said daru piyagi 💀💀
Iykyk 💀💀
Mai yahi comment dudne aya tha😂😂
oo daru badnaam kardi
Macher ka pate bhar Aaraha hai
😂
1:15 Are triggu tum samaj nahi rahe ho....wo indian tv serial ki Hulk he😂😂😂
Bhai ap aesa he hero ho yea sab kar ne ke kya jarurat hai 😊😊
11:53 Pr aap to green forest hoo🤭
5:26
the power of statue 😂
10:06 nonu bhaiya to pehle se hi hero material hai ❤
Pagle
Yeah you are right❤ nishchay bhaiya are very handsome and cute boy❤😊
Mere sbhi dekhe hue serial ko roast kr diya bhai ne 😂😂
5:09 bro is still living in childhood😭😭✋🗿
3:21 Nation wants to know 😂😂
😅 If you saw it😁 you saw it😅 (headset in girls ears )
10:16 kuch nhi bhi karoge to bhi ap hero material hi ho😂
4:49 to bhaiya comment kaise karenge😶
😂😂
😂😂
1:45 🌚daru was personal 🐸
6:25 .. when an engineer become roster 😂😂😂😂
Now I knew it ki Bhaiya uss samay boot camp per the aur vahin per shooting kar rahe the😂😂😂❤❤❤❤❤❤❤❤❤
3:57 oh God yeah kya dekh liya 😂😂