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

Question #226700
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?
1
2021-08-20T01:37:00-0400

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"

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