Question

A ball is thrown vertically up wards with a velocity of 49 m/s calculate
i. The maximum height to which it rises, the total time it takes to return to the surface of the earth.

Answer 1

(g = 9.8m/s²)

H= u²/2g

H= 49²/19.6

H= 122.5m

T= 2u/g

T= 98/9.8

T= 10s

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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