When a ball is thrown vertically upward from the ground, it crosses a point which is at a height of 25m twice in an interval of 4 seconds. At what velocity was the ball thrown?


The time interval between the object to the same point = 4 s

It is understood that at every 2-second interval, the ball reached the same height twice.

The velocity of the ball, v = u + (-gt)

0 = u – 10 x 2

u = 20

It is also given that at velocity 25m, time is 20 seconds.

Therefore, the initial velocity is, -v2 = u2 + (-2gh)

200 = u2 – 2 x 10 x 25

u2 = 900

Therefore, u = 30 m.s-1


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