Question

A boy standing on the roof of a building
44.1 m high drops a stone. After
0.5 s, he throws another stone vertically
downwards. If both the stones reach
ground level at the same time, what is the
speed of the second stone thrown by the
boy? (g = 9.8 m/s2)
​

Answer 1

Explanation:

For first stone

s = ut + 1/2at^2

=> 44.1 = 1/2×9.8×t^2 ( °•° u = 0 for first stone)

=> 44.1 = 4.9t^2

=> t^2 = 44.1/4.9

=> t = 3s

Now ,for sencond stone

time t = 3-0.5 = 2.5s

s = ut + 1/2at^2

=> 44.1 = u(2.5)+1/2×9.8×(2.5)^2

=> 44.1 = 2.5u + 4.9×6.25

=> 44.1 = 2.5u + 30.625

=> 13.475 = 2.5u

=> u = 5.39m/s

Hence, initial velocity of second stone is 5.39m/s