Answer to Question #94817 in Mechanics | Relativity for Avinnash Suthaharan

Question

A ball is thrown upward from the ground. You observe the ball through a window on its way up, and notice that it was visible for 1.8 seconds while it travels from the bottom of the window to the top, which is a length of 40.9122 metres.

How much time does it take for the ball to be seen again, in seconds?
How far above the top of the window will the ball reach, in metres?

Answer 1

Let us write the equations of motion of the ball, as if it started moving from the bottom of the window, with some initial speed $v_1$. The equations are then $y(t) = H + v_1 t - \frac{g t^2}{2}$, $v = v_1 - g t$, where $H$ is the distance from the ground to the bottom of the window. Let the length of the window and time it takes to move through it upwards be $L = 40.9122 m$ and $T = 1.8 s$ respectively.

When the ball is at the top of the window, $y(T) = H + L = H + v_1 T - \frac{ g T^2}{2}$ , from where we can find $v_1 = \left( L + \frac{g T^2}{2} \right)/T$ .

When the ball reaches the highest point, $v = 0 = v_1 - g t'$ , hence $t' = \frac{v_1}{g}$. So, the corresponding coordinate of the ball is $y(t') = H + \frac{v_1^2}{2 g}$ . Hence, the distance between the top of the window and the highest point is $y(t') - (L + H) = \frac{v_1^2}{2 g} - L = 9.85 m$ .

The time it takes to see the ball again is two times the time it takes to reach the highest point from the top of the window, that is $2 (t' - T) = 2 (\frac{v_1}{g} - T) = 2.83 s$.