Projectile Motion: Shooting over a hill

  • Thread starter kkernodl
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Homework Statement



A projectile is fired with speed v0 at an angle θ from the horizontal as shown in the figure (see attachment).

Homework Equations



A. Find the highest point in the trajectory, H. Express the highest point in terms of the magnitude of the acceleration due to gravity g, the initial velocity v0, and the angle θ.

B. What is the range of the projectile, R? Express the range in terms of v0, θ, and g.

C. Find the angle theta above the horizontal at which the projectile should be fired. Express your answer in terms of H and R.

D. What is the initial speed? Express v0 in terms of g, R, and H.

E. Find tg, the flight time of the projectile. Express the flight time in terms of H and g.

The Attempt at a Solution



A. H=(v0sin(θ))2)/2g

B. R=v02sin(2θ)/g

C. I know that arctan(2H/R)=θ, however this isn't the correct answer. I've tried everything I can think of, but don't know how to solve this part.
 

Attachments

  • Screen Shot 2012-02-29 at 12.02.35 PM.png
    Screen Shot 2012-02-29 at 12.02.35 PM.png
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Answers and Replies

  • #2
hotvette
Homework Helper
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C. I know that arctan(2H/R)=θ, however this isn't the correct answer

What you need to do is find the time it takes for the range (x) to be R/2 and then find θ such that the height (y) is H.

Your expression is valid for a triangle connecting the starting point to the peak but that doesn't have anything to do with the projectile problem.
 
  • #3
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What you need to do is find the time it takes for the range (x) to be R/2 and then find θ such that the height (y) is H.

Your expression is valid for a triangle connecting the starting point to the peak but that doesn't have anything to do with the projectile problem.

I found the position function in the x direction and set it equal to R/2. This gave me that the time, t, such that the position of x is R/2 is t=v0sin(2θ)/2gcosθ.

I can use trig identities to simplify this to t=v0sinθ/g

I can find the position function of y: Sy=v0t-4.9t2, but I'm not sure what to plug in.

I'm also very confused as to how I'm supposed to express the answer in terms of H and R, as the question specifically asks.
 
  • #4
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I was able to answer it using my original answer.

I had to find the ratio of H/R, which is sinθ/4cosθ or (1/4)tanθ.

I set (1/4)tanθ=H/R and solved for θ, which gave me θ=arctan(4H/R), which is the correct answer.
 

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