Question

a ball is thrown vertically upwards with a velocity of 49 m/s. clculate the maximum height which it is rises

Answer 1

Hi friend
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Your answer
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here,

Final velocity = V = 0 m/s ,

Initial velocity = u = 49 m/s ,

Acceleration due to gravity = a = g = - 9.8 m/s² ,

Height = s = ?

So,
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v² - u² = 2as

=> v² - u² = 2×g×s

=> (0)² - (49)² = 2 × (- 9.8) × s

=> - (49 × 49) = - ( 2 × 9.8 × s)

=> s = (49 × 49) / (2 × 9.8)

=> s = 122.5 m

HOPE IT HELPS

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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