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

distance travelled = d = vox*t t = d/vox = d/vi*costheta

ID: 1458467 • Letter: A

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 39 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.

distance travelled = d = vox*t

t = d/vox = d/vi*costheta <<---------(1)

along vertical

initial velocity voy = vi*sintheta

displacement y= h

acceleration ay = -g = -9.8 m/s^2

from equations of motion y = voy*t + 0.5*ay*t^2

h = vi*sintheta*d/(vi*costheta) - 0.5*g*d^2/(vi^2*(costheta)^2)

h = tantheta*d - 4.9*d^2/(vi^2*(cos45)^2)

part(B)

as the ball is landing vertical displacemnt = y = 0

0 = tan30*39 - (4.9*39^2)/(vi^2*(cos30)^2)

vi = 21 m/s <<---answer

part(C)

F = m*vi/t = 0.45*21/0.15 = 63 N <<---answer

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.