Question

A ball is thrown vertically upwards with an initial velocity of 49 m/s. Calculate: (i) the maximum height attained, (ii) the time taken by it before it reaches the ground again. (Take g= 9.8 m/s)

Answer 1

Initial velocity (u) = 49 m/s    Final velocity (v) = 0 m/s

Gravity in toward down = + 9.8 m/s2  Gravity in toward up = - 9.8 m/s2

(i) We know, V2 - u2 = 2gs

              or,  (o)2- (49)2 = 2 X 9.8 X S

              or,  s = - (49)2/ 2 X 9.8 = 122.5 m

Maximum height = 122.5 m

(ii) We know, v = u + gt

              or   0 = 49 +(- 9.8) X t

              or   9.8 X t =  49

              or, t = 49/ 9.8 = 5 s

If, time for upward direction = time for downward direction
Then, total time taken by the ball to return back = 5 + 5 = 10 s

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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