Question

A ball is thrown vertically upwards with a velocity of 49m/s. Calculate (i) the maximum height to which it rises, (ii) the total time it takes to return to the surface of the earth.
Please can you answer correctly with the subdivisions??

Answer 1

Hey there!!

Do find your solution below!!

Plz note that here I have taken the value of g=10m/s2 ..U may get the answer different if u take g as 9.8m/s2


# hope it helps you!!

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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