Projectile motion - velocity/time unknown

[zombiemegan22]

A batter hits a pitched ball at a height of 4 feet above the ground so that its angle of projection is 45 degrees and its horizontal range is 350 feet.
The ball is fair down the left field line where a 24 foot high fence is located 320 feet from home plate.
By how much will the ball clear the fence?

vfy=viy+at
(vf)^{2}= (vi)^{2}+2aD (..we'll say D= displacement)
D= vi(t)+.5(a)(t^{2})

Vx=Dx(t)

Well, I only really know my x displacement, my acceleration constant, and that my initial y velocity and my x velocity are the same because of the 45 degree triangle [at least that's what my teacher said]...

So, help?! It's been bugging me for days.

 

[physics girl phd]

Once you have your components of initial velocity along X and Y, X (horizontal) and Y (vertical) equations of motion are only related by time since there is no force along X and the gravitational force along Y, and no constraints on the object like a plane or a string, etc. (it's in free-fall) ... therefore:

First find the time when the ball is passing the fence.

Using this time, find the Y location (the vertical displacement).

Since the fence is only so tall, you can know if it clears and by how much.

P.S. Welcome to the forums!