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.
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
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)
6:46 Velocity of body = 20m/s Angle of projection = 30degree gravity = 9.8m/s^2 Let R be the range R = v^2.sin2θ/g :. R = 35.3469 m So the scooter can travel a distance of 35.35 m The average length of the car is 4.5 m There are 6 cars. :. total distance = 27 m let each space = 1 m :. total space = 7 m (7 spaces. Let each car be in the middle of each space) :. Total distance needed to be covered = 34 m 35.35 m > 34 m left due to air resistance the distance is reduced by 1m :. distance covered = 34.35m So from the above numerical, we can prove that the scooter can cross the jam.
6:37 at this point you asked us a ques Ans. u=20km/h=5.5m/s g=10m/s*2 Range of a projectile= u^2*sin2theta/g =5.5^2 x sin60/10 =3.025 x √3/2 =2.6165 Pls like, it took some effort to calculate ❤❤❤❤ Length of 1 car= 4.8m Length of 5 car=24m Theta=30 The range of projectile=u^2 x sin60/g 240=u^2 x √3/2 u^2 = 240 x 2/√3 u=4√30/√√3 Therefore required velocity=4√30/√√3 CHEEN TAPAK DUM DUM
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 👍
Aoa, so as we have data lets calculate vi=20km/h=5.56m/s theta=30-degree horizontal distance=? so for horizontal component v=vx=vcos@ where @=theta vx=5.56*cos(30) =4.83m/s for horizontal time t=2*vx/g t=2*4.83/9.8 ≈0.99sec for now horizontal distance d=vx*t=5.56*0.99 so d≈4.78m since i used AI to solve this answer as i didn’t study projectile motion yet Jazak Allah khair
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.
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 ❤️
6:35 lets say 18km/hr therefore 5m/s and slope 30 degree so 5 cos 30 and motion will be of 0.5s projectile so 5 * cos 30 * 0.5 = 5 root 3 / 2 *2= approx 2.125m metres!!!
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😅😅
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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.
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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
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 😂😂
6:46
Velocity of body = 20m/s
Angle of projection = 30degree
gravity = 9.8m/s^2
Let R be the range
R = v^2.sin2θ/g
:. R = 35.3469 m
So the scooter can travel a distance of 35.35 m
The average length of the car is 4.5 m
There are 6 cars.
:. total distance = 27 m
let each space = 1 m
:. total space = 7 m (7 spaces. Let each car be in the middle of each space)
:. Total distance needed to be covered = 34 m
35.35 m > 34 m
left due to air resistance the distance is reduced by 1m
:. distance covered = 34.35m
So from the above numerical, we can prove that the scooter can cross the jam.
Insane bro ,itni mehnat kyu Bhai 😅
JEE aspirants be like:
bhai 20km/hr speed ka kaha tha , 20m/s speed bohot jyada hoti hai which is 72km/hr
bakloli kar di bro wrong speed lol
@@omnimamba8611 also friction was ignored here especially when it was on the slanted surface
6:37 at this point you asked us a ques
Ans. u=20km/h=5.5m/s
g=10m/s*2
Range of a projectile= u^2*sin2theta/g
=5.5^2 x sin60/10
=3.025 x √3/2
=2.6165
Pls like, it took some effort to calculate ❤❤❤❤
Length of 1 car= 4.8m
Length of 5 car=24m
Theta=30
The range of projectile=u^2 x sin60/g
240=u^2 x √3/2
u^2 = 240 x 2/√3
u=4√30/√√3
Therefore required velocity=4√30/√√3
CHEEN TAPAK DUM DUM
😂😂
🎉🎉 we haven't started trigonometry in class yet so I won't understand the solution BTW nischay should give heart to ur comment
Thankyou guys for the appreciation
Bro kuch samajh nhi aya 😂😂 I am. Commerce student
WHAT bro you r genius btw I m only in sixth but that trigonometry should be like basic
3:48 gaming insaan set up❤
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😮
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6:44 Velocity = 20m/s, gravity = 9.8, angle of elevation =45degree
so range = v^2/g x sin2theta
= 20^2/9.8 × sin2×45°
=400/9.8 ×sin90°
=40.81 ×1
=40.81 m
😅😅😅
Mass kidhar here baba, sas ka hi vajan 100 kilo ho jayega.
Bhai velocity 20 km per hr hai
Use chng toh kar tu galat value laga rha hai yrr
(U²sin2theta)/g hota hai na
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I
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Bhai those marzil and vanjaj ke scenes gave me nostalgic memories back ❤.....
Becuz I'm re-watching this video after completing gaming insaan ❤
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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 👍
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Mind blowing sir g 🤓 konsi class mai ho aap 😭
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V=20km/hr X 5/18 =100/18m/s
Horizontal range = ux * t= 100/18* cos30 *t°= 5.55*√3/2*t
We know t=2usintheta/g
T=2*5.55*0.5/10= 0.555. (Sin30°=0.5)
Horizontal range=5.55*√3/2*t
4.8063*0.555=2.667m range
Ans=2.667m udi hai scooty se
Topic(projectile motion)
Or you can also apply R= u²sin2Q/g
🫡🫡🫡🫡🫡
Bro good math
BRO JEE ASPIRANT YA NEET ASPIRANT
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11:00 Caption gaya tel mein 😶🥲
Aoa,
so as we have data lets calculate
vi=20km/h=5.56m/s
theta=30-degree
horizontal distance=?
so
for horizontal component
v=vx=vcos@
where @=theta
vx=5.56*cos(30)
=4.83m/s
for horizontal time
t=2*vx/g
t=2*4.83/9.8
≈0.99sec
for now horizontal distance
d=vx*t=5.56*0.99
so
d≈4.78m
since i used AI to solve this answer as i didn’t study projectile motion yet Jazak Allah khair
Abhi just projectile motion padh rhi thi 😂
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Bro formula is u^2sin2◇÷g😂
Agar projectile motion padha hota to ye question 5 line ka tha
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@@RG-lo8wr I wanna know the same
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That's great
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.
6:42 Hacker hai bhai💀💀
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Tum kon ho
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11:26 he is such a green universee..😅❤
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
0:57 Shuru ho gaya roast 😂🔥 with physics
Physics m b ek kar rakhi saloo me
3:56 was epic finally nishchay ki shadi to hui chahe noni banake hui😂😂
Tum kon ho 😮
3:04 the female character is my sister 😊
Liar 😂😂
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 ❤️
0:01 reminds me of old Ashish Chanchlani videos 😂❤
Kamla OP😂😂
Sameeee😢 nostalgia ❤
Same❤
0:07 hehe
Exactly 💯😂
2:45 ap janata ko murk samjhna band kare ...hagu😂😂❤
Tum kon ho 😮
07:28 😂😂😂 ooh ohh ohho ohhoo 😂😂
12:55 haas rahe ho ya ro rahe ho 😅
12:50 the mood swings bruhhhh 😂😂😂
4:49 to bhaiya comment kaise karenge😶
😂😂
😂😂
The way at 5:30 nishchay bhaiya is wearing a t-shirt that says mistake....These indian tv serials are actually a mistake😂😂
6:40
velocity=20km/h=5.55m/s
maximum range on angle 30=((5.55)²*sin(2*30))/g
=(30.80)*1.73/2*9.8
≈ 2.718 metres(mass neglected)
air drag lele account mein
2:11 this is so funny 😂😂😂😂
9:06 sasural simar ka❌
Makkhi✔️
Makhija returns
Sasural makkhi ka😂😂
3:43 - 4:09 THE BEST PART OF THE VIDEO 😂😂😂
11:15 noughty ho raha ke 🌚
Nhi samjhi
@@Priyanka86670 Jake pogo dekh 😂
Samaj jao 🌚🌚hmmmm 🌚🌚@@Priyanka86670
Thoda meaning bolo koi 😢
@@ponnivedhad4794 Pogo dekho jake
14:12 bahubalii rootiiiii💀
6:35 lets say 18km/hr therefore 5m/s and slope 30 degree so 5 cos 30 and motion will be of 0.5s projectile so 5 * cos 30 * 0.5 = 5 root 3 / 2 *2= approx 2.125m metres!!!
Ayein 😂
Bhai time of flight 0.5 sec aayega naa
And acceleration bhi too hai let's say 5 m/s² then approx 2.7m ke aaspass 😂😂😂😂
@@ironheadgamerzachaa ha bhai thank you time of flight ke liye speed sin theta dekhna hoga which is 2.5m/s maine main speed consider kar li👍
@@ironheadgamerzafter correction it will be 5 root 3 by 4 means 8.5 by 4 so 2.125 not 2.7 i guess!
@@hetviklakhani3209 and acceleration bhi too consider karege in x direction. It's a real case not ideal 👍🏻😁
The fact that nobody is talking abt the different ppl and the room😂
* I came here after watching the epic and legendary show *
4:02 HE IS 😭😭✋
I CANT YRRR😂
He is who?
@@kamalnayan10thb42 hilarious..!!😭😭✋
@@Aryatiwari1709 He is who?
12:52 old laugh is back 😂
10:07 hayee bhaiyaa hamare hero❤
Kabhi madam sir dekho ..... Karishma Singh jabardast roasting karengi❤❤❤❤😂😂😂
10:05 i could practically smell the edits
Yeah, I know how shit smells.
Just like trash
5:16 Statue scene ft. Nonu 😂🔥
English or Spanish 😂
It's giving English or Spanish vibe 👽💀🤷😂
थैंक्स for 6 subscribor complet ❤❤🎉🎉🎉
Think you 104 Subscribers ❤
9:19 logic 😂😂 rip
Me apki sari video dekhta hu .........koi ASI video nahi hai gispar me nahi hasa 😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂😂