Question

An artillery shell of mass 30 kg has a velocity of 250 m/s vertically upward. The shell explodes into two pieces; immediately after the explosion a fragment of mass 10 kg has a velocity of 120 m/s straight downward. How high above the point of the explosion does the larger fragment rise?

Answer 1

Answer:

9654.34 m

Explanation:

from conservation of momentum

$\begin{aligned}30 \times 250 &=-10 \times 120+20 \times V \\20 V &=30 \times 250+10 * 120 \\V &=\frac{30 \times 250+10 \times 120}{20}=435 \mathrm{~m} / \mathrm{s}\end{aligned}$

And from Conservation of Energy

$\frac{1}{2} m v^{2}=m g h\\h=\frac{v^{2}}{2 g}\\h=\frac{(435(m/s))^{2}}{2 \times 9.8(m/s^{2} )}\\h=9654.34 (m)$