Question

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

Answer 1

Answer:

Time in air is 7.5 s.

Step-by-step 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.

$s = u t + 0.5 gt^2\\\\5 = 80 t - 16 t^2\\\\16 t^2 - 80 t + 5 = 0 \\\\t = \frac{80\pm\sqrt{6400 - 320}}{32}\\\\t = \frac{80\pm78}{32}\\\\t = 5 s or 0.0625 s$

So the tie is 5 s

The time in air is

T = 2.5 + 5 = 7.5 s