Question

A ball is thrown vertically upward with a velocity of 19.6 m/s. Find

(i) The maximum height reached by the ball.

(ii) The time taken by the ball to reach the maximum height.

Answer 1

at maximum height
final velocity =0

v^2=u^2+2as
0=u^2-2gH
H=u^2/2g=(19.6)^2/2(9.8)=19.6m


2) v=u+at
0=u-gt
t=u/g=19.6/9.8=2sec

Answer 2

_/\_Hello mate__here is your answer--

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v = 0 m/s and

u = 19.6 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 − 19.6^2 = 2(−9.8)ℎ

⇒ ℎ =19.6×19.6/ 2×9.8 = 19.6m

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Let t be the time taken by the ball to reach the height 19.6 m, then according to the equation of motion

v= u + gt

=>0 = 19.6+ (−9.8) t

⇒t 9.8 = 19.6

⇒ t= 19.6/9.8 =2 s

But, Time of ascent = Time of descent

Therefore, total time taken by the ball to return

= 2+ 2= 4 s

I hope, this will help you.☺

Thank you______❤

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