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 figure). 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 $$g = 10 m s^{-2}$$ and $$\sqrt{3} = 1.7$$)

Solution

## Total time of flight of one object is $$t=\dfrac { 24\sin { \theta } }{ g }$$$$=\dfrac { 24\sin { 30° } }{ g } =\dfrac { 4 }{ g }$$In this period both object travel 100m horizontal distance 50 m with speed $$4\cos 30°$$ where e = coefficient of resistution Now $$t=\dfrac { 50 }{ u\cos { 30° } } +\dfrac { 50 }{ eu\cos { 30° } }$$$$\dfrac { u }{ g } =\dfrac { 50 }{ 4\times \sqrt { \dfrac { 3 }{ 2 } } } +\dfrac { 50 }{ 0.7\times \sqrt { \dfrac { 3 }{ 2 } } 4 } \\ { u }^{ 2 }=g\left( 100\left( \dfrac { 1 }{ \sqrt { 3 } } +\dfrac { 1 }{ 0.7\times \sqrt { 3 } } \right) \right) \\ =100g\times 1.428\\ { u }^{ 2 }=1428=(37.8)^{ 2 }\\ u\approx 38.m/s$$Physics

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