A soccer player is at a park and kicks a ball with an initial speed of 18.00 m/s

Question

A soccer player is at a park and kicks a ball with an initial speed of 18.00 m/s at an angle of 40 degrees above the horizontal. There is a 5m high wall 16m away from the child and in the direction of the kick. There is negligible air resistance. (use g = 9.81m/s^2)

a) What is the magnitude of the ball's velocity at the top of the trajectory (at the max height)?

b) What is the max height of the ball in its trajectory?

c) What is the x position when the ball reaches its maximum height?

d) What is the ball's velocity just before it htis the ground?

e) How far away from the child does the ball land? f) Does the ball hit the wall or go over it?

Explanation / Answer

Initial speed = 18.00 m/s

Angle = 40 degree above horizontal

Height of wall = 5 m

Distance of wall from child = 16 m

g = 9.81 m/s2

a) the magnitude of the ball's velocity at the top of the trajectory (at the max height) = Vcos(40) = 18 Cos(40)

= 13.79 m/s

b) the max height of the ball in its trajectory

H = u2Sin2(40) / 2g = 182Sin2(40) / 2*9.81 = 6.82 m

c) the x position when the ball reaches its maximum height will be R/2 whre R is the range of the projectile given by R = u2Sin(2*40) / g = 32.53 m

R/2 = 16.27

So the x position when the ball reaches its maximum height is 16.27 m

d) the ball's velocity just before it hits the ground = 18 m/s SAme as initial

e) How far away from the child does the ball land

R = 32.53

f) Does the ball hit the wall or go over it?

The wall will go over it. Since at x= 16 m

height > 5 m

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