- #1

**[SOLVED] projectile motion problem**

Hello guys,

This particular problem has been giving me a headache. I can't see how to figure out this problem w/o being given an inital angle. Here it is:

A baseball is hit at ground level. The ball reaches its max height above ground level 3 s after being hit. Then 2.5 s after reaching its max height, the ball barely clears a fence that is 97.5 m from where it was hit. Assume the ground is level. a) What max height above ground level is reached by the ball? b) How high is the fence? c) How far beyond the fence does the ball strike the ground?

Ok, to start out I drew the picture and labeled it and everything. Then..

a)

Y-Yo = Vt - 1/2 at^2

max height = m = 0(3) - 1/2 (-9.8)(3)^2 = 44.1 m

Ok, this is where I get stuck. I'm thinking about using the trajectory equation, to get the height of the fence.

y = (tan@)x - (gx^2)/(2(Vocos@))^2

I'm thinking that y = the height of the fence and x is the distance to the fence. So, x = 97.5 m.

But I don't know how to get the angle, @!

It's been driving me nuts because I tried different ways, but kept getting answers that didn't match.

Here's some of what I tried:

X = Xo + Vt - 1/2 at^2

X = 0 - 1/2 (-9.8)(3)^2 = 44.1m

Then since X = 44.1= Y

I did (44.1^2 + 44.1^2)^(1/2) = 62.4 m

Then sin@ = 44.1/62.4 --> @ = about 45 degrees

Is that correct?

How should I approach the rest of the problem?

Thanks much!