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(D) Let the first ball be thrown at $t = 0$ and the second ball at $t = 3 \, 	ext{s}$.Let $t$ be the time elapsed since the first ball was thrown.The

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(D) Let the first ball be thrown at $t = 0$ and the second ball at $t = 3 \, 	ext{s}$.Let $t$ be the time elapsed since the first ball was thrown.The position of the first ball is $y_1 = u t - \frac{1}{2} g t^2 = 35 t - 5 t^2$.The position of the second ball is $y_2 = u(t - 3) - \frac{1}{2} g(t - 3)^2 = 35(t - 3) - 5(t - 3)^2$.At the point of collision, $y_1 = y_2$.$35 t - 5 t^2 = 35(t - 3) - 5(t^2 - 6 t + 9)$$35 t - 5 t^2 = 35 t - 105 - 5 t^2 + 30 t - 45$$0 = 30 t - 150$$30 t = 150 \implies t = 5 \, 	ext{s}$.Substituting $t = 5 \, 	ext{s}$ into the equation for $y_1$:$h = 35(5) - 5(5)^2 = 175 - 125 = 50 \, 	ext{m}$.Thus, the balls collide at a height of $50 \, 	ext{m}$.

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