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 1.9 seconds while it travels from the bottom of the window to the top, which is a length of 51.71705 metres.

(a)
How much time does it take for the ball to be seen again, in seconds?

Answer 1

Answer on Question #72469-Physics-Other

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 1.9 seconds while it travels from the bottom of the window to the top, which is a length of 51.71705 metres.

How much time does it take for the ball to be seen again, in seconds?

Solution

$$h = v t - \frac{g t^{2}}{2}.$$ $$h_1 = v t_1 - \frac{g t_1^2}{2}$$ $$h_2 = v t_2 - \frac{g t_2^2}{2}$$ $$h_2 - h_1 = v (t_2 - t_1) - \frac{g}{2} (t_2^2 - t_1^2)$$ $$v = \frac{h_2 - h_1}{t_2 - t_1} + \frac{g}{2} (t_1 + t_2)$$

At maximum height:

$$v - g T = 0, T = \frac{v}{g}.$$ $$T = \frac{1}{g} \frac{h_2 - h_1}{t_2 - t_1} + \frac{1}{2} (t_1 + t_2)$$

We need to find:

$$t' = 2 (T - t_2) = 2 \left( \frac{1}{g} \frac{h_2 - h_1}{t_2 - t_1} + \frac{1}{2} (t_1 + t_2) - t_2 \right) = \frac{2}{g} \frac{h_2 - h_1}{t_2 - t_1} - (t_2 - t_1)$$ $$t' = \frac{2}{9.81} \frac{51.71705}{1.9} - 1.9 = 3.65 \text{ s}.$$

Answer: 3.65 s.

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