Kicked Ball

## Question:

When a ball is kicked from point A it just clears a wall at point B at the ball's maximum height. It is 20 meters from A to B and the wall is 4 meters high. Determine the initial speed with which the ball was kicked, neglecting the size of the ball and air resistance.

The expression for horizontal position of the ball is x=Vh*t. The expression for vertical position of the ball is y=1/2*(-9.8)*t2+Vv*t. At the time the ball clears the wall y=4m and x=20m so to solve for the three variables Vh, Vv and t at the time the ball clears the wall, plug in those values for x and y. This gives us two equations,

a) 4=1/2*(-9.8)*t2+Vv*t and b) 20=Vh*t.

Since we need to detrmine Vh, Vv and t, we need more information. The ball clears the wall at its maximum height, meaning that c) Vv-9.8*t=0 at time=t, or Vv=9.8*t

Now let's put 9.8*t in for Vv in equation a)

4=1/2*(-9.8)t2+9.8*t2
4=-4.9t2+9.8t2=4.9*t2
t2=4/4.9
t=0.9 seconds

Now put 0.9s in for t in a) and b) to get Vv and Vh 4=-4.49*.81+.9*Vv
Vv=(4+3.64)/.9=8.49m.s

20=Vh*.9
Vh=20/.9=22.2m/s

Initial speed is the square root of the sum of Vh2 and Vv2

speed=(22.22+8.492)^.5=23.7m/s

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