Question

You throw a ball from the balcony onto the court in the basketball arena. You release the ball at a height of 6 m above the court, with an initial velocity equal to 9 m/s at 33° above the horizontal. A friend of yours, standing on the court 10 m from the point directly beneath you, waits for a period of time after you release the ball and then begins to move directly away from you at an acceleration of 3 m/s2. (She can only do this for a short period of time!) If you throw the ball in a line with her, how long after you release the ball should she wait to start running directly away from you so that she'll catch the ball exactly 1 m above the floor of the court?

Answer 1

Answer on Question#54851 - Physics - Mechanics - Kinematics - Dynamics

You throw a ball from the balcony onto the court in the basketball arena. You release the ball at a height of $H_{i} = 6\mathrm{m}$ above the court, with an initial velocity equal to $v = 9\frac{\mathrm{m}}{\mathrm{s}}$ at $\varphi = 33{}^{\circ}$ above the horizontal. A friend of yours, standing on the court $L = 10\mathrm{m}$ from the point directly beneath you, waits for a period of time after you release the ball and then begins to move directly away from you at an acceleration of $a = 3\frac{\mathrm{m}}{\mathrm{s}^2}$. (She can only do this for a short period of time!) If you throw the ball in a line with her, how long after you release the ball should she wait to start running directly away from you so that she'll catch the ball exactly $H_{f} = 1\mathrm{m}$ above the floor of the court?

Solution:

To find the time the ball spent in the air it's useful to write the dependence of the ball's height $h$ from time $t$ ($t = 0$s when the ball is released):

$$h(t) = H_{i} + v \cdot \sin \varphi \cdot t - \frac{gt^{2}}{2},$$

where $g = 9.8\frac{\mathrm{m}}{\mathrm{s}^2}$ – is the acceleration due to gravity. To find the time it spent in the air we should solve the previous equation for $h(t) = H_f$:

$$H_{f} = H_{i} + v \cdot \sin \varphi \cdot t - \frac{gt^{2}}{2}$$ $$1\mathrm{m} = 6\mathrm{m} + 9\frac{\mathrm{m}}{\mathrm{s}} \cdot \sin 33{}^{\circ} \cdot t - \frac{9.8\frac{\mathrm{m}}{\mathrm{s}^2} \cdot t^2}{2}$$ $$5\mathrm{m} + 4.9\frac{\mathrm{m}}{\mathrm{s}} \cdot t - 4.9\frac{\mathrm{m}}{\mathrm{s}^2} \cdot t^2 = 0$$

This equation has only one positive root

$$t = 1.6\mathrm{s}$$

Since the horizontal speed of the ball is constant and equal to $v_{h} = v \cdot \cos \varphi = 9\frac{\mathrm{m}}{\mathrm{s}} \cdot \cos 33{}^{\circ} = 7.5\frac{\mathrm{m}}{\mathrm{s}}$, the ball overcomes the distance

$$l_{f} = v_{h} \cdot t = 7.5\frac{\mathrm{m}}{\mathrm{s}} \cdot 1.6\mathrm{s} = 12\mathrm{m}$$

Therefore the friend should overcome the distance of $l_{f} - L = 12\mathrm{m} - 10\mathrm{m} = 2\mathrm{m}$ to catch the ball. She will need some time $\tau$ to do this. This time is given by

$$\tau = \sqrt{\frac{2(l_{f} - L)}{a}} = \sqrt{\frac{2 \cdot 2\mathrm{m}}{3\frac{\mathrm{m}}{\mathrm{s}^2}}} = 1.2\mathrm{s}$$

Therefore after the ball was released she should wait for the following time

$$t - \tau = 1.6s - 1.2s = 0.4s$$

Answer: 0.4s.

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