A 0.45-kg soccer ball is kicked at an angle above the ground with initial speed

Question

A 0.45-kg soccer ball is kicked at an angle above the ground with initial speed vi.

Part A

Determine the height of the ball as a function of the horizontal distance d it travels down the field.

Express your answer in terms of acceleration due to gravity g, some or all of the variables , vi and d.

Part B

For = 30 , what initial speed must the ball have in order to land 45 m from where it is kicked?

Express your answer with the appropriate units.

Part C

If the foot is in contact with the ball for 0.15 s, what is the average force exerted by the foot on the ball in the previous part?

Express your answer with the appropriate units.

Explanation / Answer

A)

d = vi*cos()*t

then t = d/(vi*cos())

y = (vi*sin()*t) - (0.5*g*t^2)

y= h

then h = (vi*sin()*d/(vi*cos())) - (0.5*g*(d/(vi*cos()))^2)

h = d*tan() - 0.5*g*[d/(vi*cos())]^2

B) d = vi^2*sin(2*)/g

45 = vi^2*sin(2*30)/g

45*9.81 = vi^2*sin(60)

vi = 22.57 m/s

C) Favg*dt = 0.45*2*vi*sin(30) = 0.45*2*22.57*0.5 = 10.1565

Favg = 10.1565/0.15 = 67.71 N

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