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(A) Let the ball be thrown at $t = 0$. The vertical motion of the ball is independent of the train's acceleration.For vertical motion,the displacement

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$5$

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(A) Let the ball be thrown at $t = 0$. The vertical motion of the ball is independent of the train's acceleration.For vertical motion,the displacement $s_y = 0$ when the ball returns to the initial height.Using $s_y = u_y t - \frac{1}{2} g t^2$,where $u_y = 10 \sin 60^{\circ} = 10 	imes \frac{\sqrt{3}}{2} = 5\sqrt{3} \ m/s$ and $g = 10 \ m/s^2$:$0 = 5\sqrt{3} t - 5 t^2 \implies t = \sqrt{3} \ s$.In the frame of the train,the ball has a horizontal acceleration of $-a$ (opposite to the train's acceleration).The horizontal displacement of the ball relative to the boy is $s_x = u_x t - \frac{1}{2} a t^2$,where $u_x = 10 \cos 60^{\circ} = 5 \ m/s$.Given $s_x = 1.15 \ m$ and $t = \sqrt{3} \ s$:$1.15 = 5(\sqrt{3}) - \frac{1}{2} a (\sqrt{3})^2$$1.15 = 5(1.732) - 1.5 a$$1.15 = 8.66 - 1.5 a$$1.5 a = 8.66 - 1.15 = 7.51$$a = \frac{7.51}{1.5} \approx 5 \ m/s^2$.

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