Solution:

At the highest point, velocity is horizontal: \(v_x = u\cos(37^\circ) = 100 \times 0.8 = 80\text{ m/s}\). By conservation of linear momentum along the horizontal: \(m v_x = m_1 v_1 + m_2 v_2\). Here \(m_1 = m/4\) (comes to rest, \(v_1 = 0\)) and \(m_2 = 3m/4\). Thus, \(m(80) = \frac{3m}{4} v_2\), which gives \(v_2 = \frac{320}{3}\text{ m/s}\).