Two identical smooth balls are projected from points O and A on the horizontal ground with same speed of projection. The angle of projection in each case is 37º (see figure). The distance between O and A is 160 m. The balls collide in mid air and return to their respective points of projection. If the coefficient of restitution is 0.5, find the speed of projection of either ball (in m/s).

Engineering

Physics

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v_y direction does not change on collision. By symmetry, vx is same for each.

$e = 0 .5 = \frac{v_{x} + v_{x}}{ucos 37^{\circ} + ucos 37^{\circ}}$

v_x = 0.5u cos 37°

$t_{1} = \frac{80}{ucos 37^{\circ}}$

$t_{2} = \frac{80}{0 .5 ucos 37^{\circ}}$

$t_{1} + t_{2} = T = \frac{2 usin 37^{\circ}}{g}$

$\frac{80}{ucos 37^{\circ}} [1 + 2] = \frac{2 usin 37^{\circ}}{g}$

$u^{2} = \frac{80 \times 3 \times 10}{\frac{4}{5} \times 2 \times \frac{3}{5}} = 2500$

u = 50 m/s