Answer to Question #137732 in Physics for Key

Question

A projectile lands 250 m from the base of a vertical cliff 150 m high. How fast was it launched?

Answer 1

The distance travelled by the object in the vertical direction is:

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

If the height is $h = 150m$ and $g = 9.81m/s^2$ is the gravitational acceleration, then the falling time will be:

$$t_{fall} = \sqrt{\dfrac{2h}{g}}$$

Object's horizontal speed has not been changed with time. Thus, the distance in the horizontal direction travelled in time of falling will be:

$$d = v_0t_{fall} = v_0\sqrt{\dfrac{2h}{g}}$$

where $v_0$ is the initial velocity. If $d = 250m$, then the initial velocity is:

$$v_0 = d\sqrt{\dfrac{g}{2h}} = 250\sqrt{\dfrac{9.81}{2\cdot 150}}\approx 45.2m/s$$

Answer. 45.2 m/s.