Question

# 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 $$30^o$$(See Fig,). The distance between O and A is $$100$$m. The balls collide in mid-air and return to their respective points of projection. If the coefficient of restitution is $$0.7$$, find the speed of projection of either ball(in m/s) correct to nearest integer.(Take $$g=10 ms^{-2}$$ and $$\sqrt{3}=1.7$$).

Solution

## The collision does not alter the vertical component of the velocity of either ball so, $$t_1 + t_2 = T$$$$\dfrac{50}{ucos30} + \dfrac{50}{0.7ucos30} =\dfrac{2usin30}{10} \\ \dfrac{50 \times 2}{u\sqrt{3}} + \dfrac{50\times 2}{0.7u\sqrt{3}} = \dfrac{2u\times 0.5}{10} \\ \dfrac{70+100}{0.7u\sqrt{3}} = \dfrac{u}{10} \\ u^2 = \dfrac{170\times 10}{0.7\times 1.7} \\ u =\sqrt{\dfrac{1000}{0.7}} \\ u\approx 38m/s$$Physics

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