Question

A ball of mass 0.2 kg rests on a vertical of height 5 m. A bullet of mass 0.01 kg, travelling with a velocity V m/s in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 m and the bullet at a distance of 100 m from the foot of the post. The initial velocity V of the bullet is

Answer 1

Time to fall = t

$\frac{1}{2}g{t}^2=5$

t = 1 sec

Now
Horizontal velocity x time of flight = Range

$V_{ball}\times 1=20;V_{ball}=20m/s$

$V_{bullet}\times 1=100;V_{bullet}=100m/s$

$m_{bullet}V=m_{bullet}.V_{bullet}+m_{ball}.V_{ball}$

$0.01v=0.01\times 100+0.2\times 20$

$0.01v=1+4=5$

$v=500m/s$