How high will the projectile be at 5 km downrange?

[jdawg]

Homework Statement

A projectile is fired with an initial speed of 500 m/s at an angle of elevation of 45°. When and how far away will the projectile strike? How high overhead will the projectile be when it is 5 km downrange?

Homework Equations

The Attempt at a Solution

Flight time: t=$\frac{2(500)sin45°}{9.8}$
t=72.2 sec
How far: x=500cos45°(72.2)
x=25526.6 m

I'm not entirely sure how to figure out the second part though. I was thinking about using this formula: r(t)=v_ocosθ(t)+(v_osinθ(t)-$\frac{1}{2}$gt²)
r(t)=x+(y-$\frac{1}{2}$gt²)
Am I on the right track with this? Do I use the t I calculated earlier for this equation?

 

[haruspex]

How long will it take to be that far down range?