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During the kickoff at a local high school football game, a football is placed onto a tee to a height of 0.125 feet above the ground. The football is kicked with an initial velocity of 80 feet per second. If the ball is caught at a height of 5 ft, approximately how long did it remain in the air? Round to the nearest hundredth of a second. Answer the question and make an equation

User Wankata
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Answer:

Time in air is 7.5 s.

Explanation:

initial height, h = 0.125 feet

initial velocity, u = 80 ft/s

final height, h' = 5 ft

acceleration due to gravity, g = 32 ft/s^2

Let the time is t to reach the maximum height.

v = u - gt

0 = 80 - 32 t

t = 2.5 s

Let the time is t' from top to ground.


s = u t + 0.5 gt^2\\\\5 = 80 t - 16 t^2\\\\16 t^2 - 80 t + 5 = 0 \\\\t = (80\pm√(6400 - 320))/(32)\\\\t = (80\pm78)/(32)\\\\t = 5 s or 0.0625 s

So the tie is 5 s

The time in air is

T = 2.5 + 5 = 7.5 s

User Aabiro
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