Physics
posted by Leah .
A baseball is hit at ground level. The ball is observed to reach its maximum height above ground level 3.0 s after being hit. And 2.5 s after reaching this maximum height, the ball is observed to barely clear a fence that is 97.5 m from where it was hit. How high is the fence?

Horizontal:
97.5=vi*5.5
so vihorizontal= 97.5/5.5 m/s
vertical: at the top, the vertical veloicty is zero, so
0=viy1/2 g*3^3
viy=4.9*9g
so now we have the vertical and horizontla components.
finally
hf=viy^t1/2 g t^2 so at t=5.5, knowing viy, calculate hf. 
5.5

First, you want to find the maximum height.
At the maximum height, you know the velocity must be 0.
vi = initial velocity
vy(t) = vertical velocity
t1 = first time (3 s)
SO:
vy(t1) = g*t1 + vi*sinθ
0 = g*t1 + vi*sinθ
vi*sinθ = g*t1
Now, it's just a matter of substituting that into the equation for vertical distance.
ymax = maximum height reached by ball
y(t) = vertical distance
ymax = y(t1) = 1/2*g*t1^2 + vi*sinθ*t1
= 1/2*g*t1^2 + g*t1*t1
= 1/2*g*t1^2 + g*t1^2
= 1/2*g*t1^2
= 1/2(9.8)3^2 = 44.1m
From there, it's a simple matter of finding vi*sinθ.
44.1 = 1/2*g*t1^2 + vi*sinθ*t1
44.1 = 1/2(9.8)3^2 + vi*sinθ(3)
vi*sinθ = 29.4
When t = 5.5
y(5.5) = 1/2(9.8)5^2 + 29.4(5.5)
= 13.475m
Thus, in order for the ball to clear the fence, it must be less than 13.475 m!
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