A ball is thrown vertically upwards with a velocity of 98ms-1 calculate the maximum height to which it rises and the total time it take to return to the surface of the earth
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0
distance= 50 metres
time = 3.1945 seconds
time = 3.1945 seconds
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_/\_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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