A cannon ball is fired vertically upward with an initial speed of Vo. At the highest point it explodes into two pieces of equal masses. One of the pieces hits the ground with a speed of 2Vo.

What is the speed of the second piece when it hits the ground?

How long was the second piece in the air?

###### Solution. Posted on Wednesday, January 21, 2015.

At the highest point, the cannon ball’s velocity is zero and the total momentum of a system is zero as well. According the law of conservation of momentum

0 = mV1 + mV2 or V1 = -V2

The height of the highest point can be find from the formula

H = Vo^2/(2g)

Based on the law of conservation of energy, we can get formula for the first piece of the cannon ball

mV1^2/2 + mgH = m(2Vo)^2/2

Thereby,

V1 = Vo*√3 and V2 = -V1 = -Vo*√3

According the law of conservation of energy, we could say that the speed of the second piece when it hits the ground will be equal to the speed of the first piece 2Vo. (Since they fell from the same height with the same initial speed.)

Let’s calculate how long was the second piece in the air. The total time is the sum of t1 – the time during the upward motion and t2 – time during the downward motion.

t1 = Vo/g

t2 = √(2*H/g) = Vo/g

t1 = t2 (The time does not depend on mass.)

Finally, t = t1 + t2 = 2Vo/g .