Oliver punts a football into the air. The football has an initial velocity of 21

Question

Oliver punts a football into the air. The football has an initial velocity of 21.2 m/s and is projected at an angle of 45 deg with respect to the horizontal when it leaves Oliver’s foot. His foot is 1 m off the ground when it makes contact with the ball.

What is the maximum height reached by the football (relative to the ground)?

How long does the ball take to reach maximum height?

How long does it take the ball to fall from the maximum height to the ground?

What is the horizontal displacement of the ball when it lands on the ground (relative to its

release point)?

What is the vertical velocity of the ball just before it lands on the ground?

Explanation / Answer

a] At the maximum height, v = 0

so,

02 = uy2 - 2gh

=> h = (21.2sin45)2/(2 x 9.8) = 11.465 m

Relative to the ground, this height will be:

H = 1 + 11.465 = 12.465 m.

b] time taken to reach the maximum height will be:

t =uy /g = 21.2sin45/9.8 = 1.5296 s

c] uy = 0 [at the highest point]

H = 12.465 m

Therefore,

12.465 = 0 + (1/2)gt2

=> t = 1.595 s

this is the time it takes for the ball to reach ground from the maximum height.

c] Total travel time = T= 1.595 + 1.5296 = 3.1246 s

in this time, the horizontal displacement will be:

D = uxT = 21.2cos45 x 3.1246 = 46.84 m.

d] Vertical velocity just before the ball reaches the ground is:

vy = [2 x 9.8 x 12.465]1/2 = 15.630 m/s.

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