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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=[itex]\frac{2(500)sin45°}{9.8}[/itex]

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

_{o}cosθ(t)+(v

_{o}sinθ(t)-[itex]\frac{1}{2}[/itex]gt

^{2})

r(t)=x+(y-[itex]\frac{1}{2}[/itex]gt

^{2})

Am I on the right track with this? Do I use the t I calculated earlier for this equation?

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