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.
## Answer:

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)*t^{2}+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)*t^{2}+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)t^{2}+9.8*t^{2}

4=-4.9t^{2}+9.8t^{2}=4.9*t^{2}

t^{2}=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 Vh^{2}
and Vv^{2}

speed=(22.2^{2}+8.49^{2})^.5=23.7m/s

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