Question

A ball is thrown vertically upwards with a velocity of 98 ms 1. Calculate the maximum height to which it rises and the total time it takes to return to the surface of the earth.

CBSE Class IX Science SA 2 (3 Marks)

Answer 1

   As we know,            v² = u²-- 2gh
                               (0)² = u² – 2gh
                               u² = 2gh
                             (98)² = 2 x 9.8 x h
                             98 x 98 ÷ 2 x 9.8 = h
                               h = 98 x 98 x10 ÷ 2 x 98
                               h = 98 x 5
                               h = 490 m

 As we know,               v = u + gt
                                 (0)² = 98 + (– 9.80) x t
                                   98 = 9.8 t
                                   t = 98 ÷ 9.8
                                   t = 10 sec.
                           Total time = time to go up + time to go down
                                   = 10 + 10 = 20 sec

Answer 2

_/\_Hello mate__here is your answer--

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v = 0 m/s and

u = 98 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 − 98^2 = 2(−9.8)ℎ

⇒ ℎ =98×98/ 2×9.8 = 490 m

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Let t be the time taken by the ball to reach the height 245 m, then according to the equation of motion

v= u + gt

=>0 = 98 + (−9.8) t

⇒t 9.8 = 98

⇒ t= 98/9.8 =10s

But, Time of ascent = Time of descent

Therefore, total time taken by the ball to return

= 10 + 10 = 20s

I hope, this will help you.☺

Thank you______❤

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