Initial velocity of ball u = 30 secm
gravitational acceleration g = 1.60 sec2m
using equation motion
The maximum height reached by ball
v2−u2=−2gh
02−302=−2×1.60×h ( Final velocity v = 0, Negative sign is because of deceleration or oppose to gravity )
h= 281.25 m
The time taken to reach that height.
v=u−gt
0=30−1.60×t
t= 18.75 sec
Its velocity 20s after it is thrown.
v=u−gt
v=30−1.60×20
v=−2secm
When the ball’s height is 50m
v2−u2=−2gh
v2−302=−2×1.60×50
v=27.20secm
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