Posted by **Leah** on Sunday, May 13, 2012 at 5:48pm.

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?

- Physics -
**bobpursley**, Sunday, May 13, 2012 at 7:16pm
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=viy-1/2 g*3^3

viy=4.9*9g

so now we have the vertical and horizontla components.

finally

hf=viy^t-1/2 g t^2 so at t=5.5, knowing viy, calculate hf.

- Physics -
**abebe**, Friday, June 15, 2012 at 11:56am
5.5

- Physics -
**Sabrina**, Thursday, September 8, 2016 at 7:29pm
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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