Question

A 0.38-kg skeet (clay target) is fired at an angle of 28∘ to the horizon with a speed of 25 m/s. When it reaches the maximum height, h, it is hit from below by a 28-g pellet traveling vertically upward at a speed of 230 m/s. The pellet is embedded in the skeet.
How much higher, h′, did the skeet go up?
Express your answer to two significant figures and include the appropriate units.

Answer 1

$$h=\frac{v^2\sin^2{28}}{2g}=\frac{25^2\sin^2{28}}{2(9.8)}=7 m.$$

From the conservation of momentum

$$mu=(m+M)v'$$

From the conservation of energy:

$$0.5(m+M)v'^2=(m+M)gH$$

$$H=\frac{v'^2}{2g}=\frac{u^2m^2}{2(m+M)^2g}=\frac{230^2(0.028)^2}{2(0.028+0.38)^2 (9.8)}=13 m.$$

So,

$$h'=h+H=7+13=20 m.$$