Question

A police officer chases a thief across city rooftops. They are both running when they come to a
gap between buildings that is 4.00 m wide and has a drop of 3.00 m. The thief, having studied a
little physics, leaps at 5.00 m/s, at an angle of 45.0° above the horizontal, and clears the gap
easily. The police officer did not study physics and thinks he should maximize his horizontal
velocity, so he leaps horizontally at 5.00 m/s.
(a) Does the policy officer clear the gap?
(b) By how much does the thief clear the gap?

Answer 1

Calculate the time the officer will fall down:

$$t=\sqrt{\frac{2h}{g}}=\sqrt{\frac{2\cdot3}{9.8}}=0.782\text{ s}.$$

How far along the horizontal will the officer jump if he left the taller building at 5 m/s horizontally?

$$R_o=vt=5\cdot0.782=3.91\text{ m}.$$

We are sorry, officer. Thank you for your service.

Make calculations for the thief. The maximum height above the building where the jump started:

$$h_\text{max}=\frac{v^2\text{ sin}^2\theta}{2g}.$$

And reaching the maximum height will take a time of

$$t_\text{up}=\frac{v\text{ sin}\theta}{g}.$$

From the point of maximum height, to land, the thief will need to fall 3 meters more:

$$H=3+h_\text{max}.$$

Falling from this height will take a time of

$$t_\text{down}=\sqrt{\frac{2H}{g}}=\sqrt{\frac{2\big(3+\frac{v^2\text{ sin}^2\theta}{2g}\big)}{g}}.$$

The range will be equal to the horizontal speed times total time:

$$R_t=v\text{ cos}\theta(t_\text{up}+t_\text{down}),\\\space\\ R_t=v\text{ cos}\theta\bigg(\frac{v\text{ sin}\theta}{g}+\sqrt{\frac{2\big(3+\frac{v^2\text{ sin}^2\theta}{2g}\big)}{g}}\bigg),\\\space\\ R_t=4.32\text{ m}.$$

The thief clears the gap for

$$\Delta R_t=R_t-G=4.32-4=0.32\text{ m}.$$