Answer to Question #226700 in Civil and Environmental Engineering for Rexie

Question

An object is released from height of 72 feet and is given an initial horizontal velocity of 22 ft/s. When the object strikes the ground, how far has it travelled in the horizontal direction?

Answer 1

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

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

If the height is $h = 72\space feet,$ and $g = 29.5\space ft/s^2$ 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 $v_0 = 22ft/s$. Thus, obtain:

$$d = v_0\sqrt{\dfrac{2h}{g}} = 22\cdot \sqrt{\dfrac{2\cdot 72}{29.5}} \approx 48.6\space feet$$

Answer. 48.6 feet.