Question

Imagine a flying super-dog that is able to accelerate horizontally and vertically. You are going to play fetch with this flying dog. Imagine throwing a rock up and forwards so that it follows the usual parabolic path away from you. As you throw the rock, the dog is waiting a few meters behind you. What is the acceleration vector for the dog if it flies in a straight line to catch the rock just as it passes the highest point on the parabola?

Answer 1

Question #81570, Physics / Other

Imagine a flying super-dog that is able to accelerate horizontally and vertically. You are going to play fetch with this flying dog. Imagine throwing a rock up and forwards so that it follows the usual parabolic path away from you. As you throw the rock, the dog is waiting a few meters behind you. What is the acceleration vector for the dog if it flies in a straight line to catch the rock just as it passes the highest point on the parabola?

Solution

1) Consider the rock as projectile. At the highest point on the parabola:

$$x = v \cos \alpha t$$ $$y = v \sin \alpha t - \frac{g t^{2}}{2}.$$ $$t = \frac{2 v \sin \alpha}{g}$$

2)

$$x = \frac{a_{x} t^{2}}{2}$$ $$y = \frac{a_{y} t^{2}}{2} - \frac{g t^{2}}{2}.$$

Thus,

$$\frac{a_{x} t^{2}}{2} = v \cos \alpha t$$ $$a_{x} = \frac{2 v \cos \alpha}{t} = \frac{2 v \cos \alpha}{\frac{2 v \sin \alpha}{g}} = g \cot \alpha$$ $$v \sin \alpha t - \frac{g t^{2}}{2} = \frac{a_{y} t^{2}}{2} - \frac{g t^{2}}{2}.$$ $$a_{y} = \frac{2 v \sin \alpha}{t} = \frac{2 v \sin \alpha}{\frac{2 v \sin \alpha}{g}} = g$$

Thus, the acceleration vector for the dog if it flies in a straight line to catch the rock just as it passes the highest point on the parabola is

$$\boldsymbol{a} = (g \cot \alpha, g)$$

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