Question

A boy throws a ball from a height of 49 m building after 1 second he throws a second bowler if both ball reaches at earth together then at what velocity he

Answer 1

Hi there,

Let the value of g = 9.8 m/s

for first Ball;

S = ut+ 1/2 gt²

49 = 0 t + 1/2 ×9.8 t²

t²= 10

t= √10 = 3.16 sec

The second ball travelled after 1 sec then 3.16- 1 = 2.16 sec

Now again,

S= ut+ 1/2 gt²

49 = u×2.16 + 9.8×2.16²/2

49 = 2.16×u + 22.86

26.14/2.16 = u

u = 12.10 m/s

Hope it will help you;