Question

A ball thrown vertically upward with a velocity of 49meter per sec then calculate the maximum height to which it rise And the total time it takes to return to the surface of the Earth
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Answer 1

Here you go.
As given initial velocity=49m/s.
acceleration=-g=-9.8m/s^2 as it is in opposite direction of velocity.
At the maximum height velocity=0m/s which is the final velocity.

So according to the third equation of motion,
V^2=U^2+2as.
Hence,S=(V^2-U^2)/2g=122.5 meter.

So now let's talk about the second question means time required to return.
For this case u=0m/s
a=g=9.8m/s^2
s=122.5m

Now apply equation,
S=ut+1/2gt^2
so t=25 second.

Thanks.
Tripathy

Answer 2

_/\_Hello mate__here is your answer--

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v = 0 m/s and

u = 49 m/s

During upward motion, g = − 9.8 m s^−2

Let h be the maximum height attained by the ball.

using

v^2 − u^2 = 2h

⇒ 0^2 − 49^2 = 2(−9.8)ℎ

⇒ ℎ =49×49/ 2×9.8 = 122.5

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Let t be the time taken by the ball to reach the height 122.5 m, then according to the equation of motion

v= u + gt

=>0 = 49 + (−9.8) t

⇒t 9.8 = 49

⇒ t= 49/9.8 = 5 s

But, Time of ascent = Time of descent

Therefore, total time taken by the ball to return

= 5 + 5 = 10

I hope, this will help you.☺

Thank you______❤

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