Question

A 100 kg gun fires a ball of 1 kg horizontally from a cliff of height 500 m. It falls on the ground at a distance of 400 m from the bottom of the cliff. Find the recoil velocity of the gun. (acceleration due to gravity=10 ms^-2)

Answer 1

Applying 2^nd kinematic equation of motion s = ut + \(\frac{1}{2}\)at²

If, u = 0, g = \(\frac{m}{s^2}\), s = 500 m or t = 10 sec.

Horizontal range = u × 10

400 = u × 10 or u = 40\(\frac{m}{s}\)

From law of conservation of momentum,

m_bu_b + m_ev_g = m_pv_b + m_Gv_G

m_b × 0 + m_G × 0 = 1 × 40 + 100v_G

100v_G = −40

Or v_G = 0.4 m/s