Projectile motion question?

Question

A football player kicks a football that has a “hang time” of 4.5s and is caught 46m away. If the ball leaves the players foot at a height of 1.5m and is caught overhead at a height of 2.1m, find the ball’s initial velocity.

Answer 1

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s=(uy x t)-1/2gt^2 +ho

Where ho= 1.5m and s = 2.1m

2.1=4.5uy- g(4.5^2)/2 +1.5

(2.1-1.5) = 4.5uy- (20.25 x 9.81)/2

0.6 = 4.5uy -99.32

99.32 +0.6 = 4.5uy

uy=99.92/4.5

uy= 22.2 m/s

uy is the vertical component of velocity.

ux is the horizontal component that does not change and is given by ux= 46/4.5= 10.22 m/s

u = squ root of [22.2^2 + 10.22^2]

u = squ root of [598.17]

u =24.4 m/s

Answer 2

Horizontal component of the initial velocity is

(46m)/(4.5s) = 10.2222... m/s.

For the vertical component of motion:

1.5 = 2.1 + vy*4.5 - (1/2)(9.8)(4.5^2) =>

vy*4.5 = -0.6 + 99.225 = 98.625 =>

vy = 21.916666...

So the magnitude of the initial velocity is

sqrt(21.916666...^2 + 10.2222...^2)

= 24.2 m/s.

• Ricky2 years agoReport

  Thank you so much for your answer.
  Could you explain more details about the part you wrote "For the vertical component of motion:
  1.5 = 2.1 + vy*4.5 - (1/2)(9.8)(4.5^2)" ? and also Why is 2.1 used in here?
  
  Thanks.