Flight time and height

  • Thread starter jdawg
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  • #1
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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)=vocosθ(t)+(vosinθ(t)-[itex]\frac{1}{2}[/itex]gt2)
r(t)=x+(y-[itex]\frac{1}{2}[/itex]gt2)
Am I on the right track with this? Do I use the t I calculated earlier for this equation?
 
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Answers and Replies

  • #2
haruspex
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How long will it take to be that far down range?
 

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