How high will the projectile be at 5 km downrange?

AI Thread Summary
A projectile is launched at 500 m/s at a 45° angle, with a calculated flight time of approximately 72.2 seconds. The horizontal distance covered by the projectile is determined to be around 25,526.6 meters. To find the height of the projectile when it is 5 km downrange, the user considers using a specific formula involving initial velocities and gravitational effects. Clarification is sought on whether to use the previously calculated time for this calculation. The discussion focuses on determining the projectile's height at a specific distance and the appropriate equations to use.
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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)=vocosθ(t)+(vosinθ(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?
 
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