Question

a man throws balls with the same speed vertically upward one after the other at an interval of 2 second .what should be the speed of the throw so that more than two ball are in the sky at any time?(g=9.8)​

Answer 1

Explanation:

given, u=5m/s,a=-10m/s^2,

[negative sign is due to downward direction]

v=0 [at Max. height velocity is zero]

height attained,

S=?

time taken, t=?

(1) from third eq. of motion

v^2=u^2+2as

0=(5)^2+2x(-10)s

s=25/20=1.25m

height attained =1.25m

(2)from first eq. of motion,

v=u+at

0=5+(-10)t

t=5/10=0.5s

time taken =0.5s