Answer to Question #136185 in Physics for george

Question

11) A man on top of a building 30 m high drops a ball. At the same time a man on the ground directly beneath the ball fires a gun. What initial velocity must the bullet have in order to hit the ball 2 m below the top of the building?

Answer 1

In time $t$ a free falling ball covers a distance $h$:

$$h = v_0t + \dfrac{gt^2}{2}$$

where $g = 9.81m/s^2$ is the gravitational acceleration and $v_0 = 0$ (no starting velocity).

Thus, the ball will cover $h_1 = 2m$ in the following time:

$$t_{meet} = \sqrt{\dfrac{2h_1}{g}}$$

The distance covered by the bullet in time $t$ is:

$$h = v_0t - \dfrac{gt^2}{2}$$

where $v_0$ is bullet's initial velocity. In time $t_{meet}$ the bullet should cover the distnace $h_2$. Thus:

$$h_2 = v_0t_{meet} - \dfrac{gt_{meet}^2}{2}$$

Then, the initial velocity is:

$$v_0 = \dfrac{h_2}{t_{meet}} + \dfrac{gt_{meet}}{2}$$

Substituting $t_{meet}$, obtain:

$$v_0 = h_2\sqrt{\dfrac{g}{2h_1}} + \sqrt{\dfrac{gh_1}{2}}$$

In numbers:

$$v_0 = 28\cdot \sqrt{\dfrac{9.81}{2\cdot 2}} + \sqrt{\dfrac{9.81\cdot 2}{2}}\approx 46.98m/s$$

Answer. 46.98 m/s.