a boy throws a ball upward and after 8 seconds he catch it again then find 1. by which velocity ball was thrown upward. 2. at what height the velocity of the ball will be zero (g= 9.8 m/s²)
Answers
Explanation:
A boy threw a ball vertically upwards. It returned to his hands after 9.8 s. What was the ball's initial speed? With what speed did it return to the boys hands? How high did it go?
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5 Answers

Satya Parkash Sud, M.Sc. Physics & Nuclear Physics, University of Delhi (1962)
Answered September 26, 2017 · Author has 3.6K answers and 9.4M answer views
The ball returned to the boy after 9.8 s of it being thrown up. Let us take the point from where the ball is thrown up as the origin and upward direction as positive and downward direction as negative.
The ball takes 4.9 s to reach the highest point and 4.9 s to return from the highest point to reach the boy. Let the initial velocity of the ball with which it is thrown up be u m/s. The ball rises for 4.9 s and it rises till the velocity of the ball becomes zero after 4.9s under the opposing action of the acceleration due to gravity. Using the relation ;
v = u + a t. We have v = 0 m/s, g=- 9.8 m/s², t= 4.9s. Substituting these values in the equation we get;
0 = u - (9.8 m/s²)× 4.9 s ;==> u= 48.02 m/s.—-(1)
The initial speed with which the ball was thrown up= 48.02m/s.
The highest point reached by the ball can be obtained by using any of the relation;
v² - u² = 2 g h.
At the highest point v= 0 m/s, u = 48.02 m/s,
g = - 9.8 m/s² and h =?
0² - 48.02²= 2×(-9.8) h; ==> h= 48.02²/2×9.8= 117.649 m.
The highest point reached = 117.649 m.
After reaching the highest point, the ball starts falling towards the boy, with initial velocity u= 0 m/s ( at the highest point its velocity had becomes zero), and an acceleration g= + 9.8m/s. Its velocity after 4.9 s of free fall is,
v = 0 m/s + 9.8 m/s² × 4.9 s = 48.02 m/s
We could have obtained the same value by equating the potential energy at 117.649 m with the kinetic energy.
m g h = ½ m v²; ==> v =( 2 g h)^½=( 2×9.8×117.649)^½=48.02 m/s.
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