Projectile PHYSICS!!!

What's The MAXIMUM HEIGHT Of The Basketball?!? #Physics #Engineering #Projectile #Motion #NicholasGKK #Shorts

[nikolaninkov]

Time isn’t needed
You can solve this energetically:
mv^2/2=mgh, h=v^2/2g
You could also derive this using
at=v0, since velocity at the end is 0

[moskthinks]

Note that time isn't needed to find the max height. At max h., vertical velocity (v) = 0, so using v^2 = u^2 + 2as will yield the same value of 20.4 m.

[ertugrulghazi]

Simply,
take Y- direction
For max height,
Final vel, V =0
Initial vel, U = 20m/s
Acceleration, a = -g = -9.8
Put, v^2-u^2 = 2as,
Ans: where s = displacement in y direction= h= 20.4

[edotrin]

if you get gravity as 10m/s there is an easy way to find out the highest possible height. it goes with 5-15-25-35-45 per sec and if you wanna get the 2 seconds of it, just 5+15=20

[jeroincababat]

I used the formula to get the maximum height of the ball by using this formula:

Ymax = Vo²/2g

Where:
Vo = initial velocity
g = acceleration due to gravity

[chaoticstarfish]

If G ≈ 10m/s², then an object at 20m/s vertically will have its speed reduced to 0m/s in approximately 2s. The average speed becomes 10m/s; multiply that by the time (2s), and the result is approximately 20m. Writing it down makes it seem so overly complicated though...

[Vinnan_Thamizh]

This formula is used as there is always gravitational influence on objects on and close to the Earth, pulling them towards the centre. If there's no gravity, then the answer would be 40m.

[spencer]

You could also learn the formula
H(maximum)= U^2(initial speed in y axis)/(2g)
If angle is given
H=U^2*sin^2(angle)/(2g).

[atharvabhardwaj]

And one more formulae can help a lot is
[U(y dirtection)]^2/2g

[atharvabhardwaj]

Bro this was easy answer is optionB
We being a jee aspirants MUST solve these question in
Iike 20-25 seconds
So yeah bit habitual with these type of ques.
😊

[yassinfarouk]

Why is projectile motion displayed on a flat plane

[JamesVonPollarca]

Thankyou absolutely i dont understand with my mind

[reiniljayjamero]

or u could use the 0^2=20^2-2(9.81)(h) to find the max height since final v in max height is 0

[kit-iu]

i used the formula V²sin²(angle) all over 2(gravity)

to get V
V = √(20)²+(34)²
angle = arctan (20/34)

then plug in to the formula

(root(20^2 + 34^2)^2 sin(arctan(20/34))^2)/(2 (9.81))

[kyrucial]

I tried solving it using H=1/2g(t)² and got approx 20m
H=1/2(9.8)(2)²
H=1/2(9.8)4
H=1/2(39.2)
H=19.something

[eu_sinan]

This can also find by the equation
H = (u²sin²θ)/2g
Here we just want to replace the u²sin²θ by square of 20 m/s since it was the vertical or y component

[dearcath]

Why aren’t we considering the horizontal velocity? Or does it not matter to find out the maximum height?

[tatelangdon]

nah this is acc so easy, look the total time 2 to the highest point is after 1 second therefore 1 second after 20m/s upward is 20 meter duh... am Chinese to trust me

[yousufshaikh]

bro please make video on Jee advance questions

[Eduardo-tqsk]

Profe, why did you write the 34m/s I the horizontal velocity?why is that info, please.