Help please! Tricky-Finding the angle reqd. to hit a target

  • Thread starter kwightman
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  • #1
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Homework Statement



A gun shoots a bullet with a velocity of magnitude 490 m/s. The goal is to hit the target 980 meters away. How high above the target must you aim to correct for gravity? (assume gun and target are same height)



Homework Equations



I worked the problem by first getting the x- and y- component equations for position and velocity. I got the x-position equation at final time T, 980 meters=490cos(theta)*T, solved for T, plugged it into the y-position equation at T, 0=-1/2(9.8)T^2 + 490sin(theta)T.



The Attempt at a Solution



Solving for theta, I got about .115 degrees as the answer. The Book says the answer is 1.15 degrees. What did I do wrong? Thank you!

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
LowlyPion
Homework Helper
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Homework Statement



A gun shoots a bullet with a velocity of magnitude 490 m/s. The goal is to hit the target 980 meters away. How high above the target must you aim to correct for gravity? (assume gun and target are same height)

Homework Equations



I worked the problem by first getting the x- and y- component equations for position and velocity. I got the x-position equation at final time T, 980 meters=490cos(theta)*T, solved for T, plugged it into the y-position equation at T, 0=-1/2(9.8)T^2 + 490sin(theta)T.



The Attempt at a Solution



Solving for theta, I got about .115 degrees as the answer. The Book says the answer is 1.15 degrees. What did I do wrong? Thank you!

I think your equations look right. I think they made an error in the final answer. Perhaps they made a simplifying approximation?

Vx = V*Cosθ =490 Cosθ
X = V*t = 980 = 490*Cosθ *t
t = 2/Cosθ

For your Y equation you get that Vy*t = 1/2 g*t2

This yields Vy = V*Sinθ = 1/2 g*t = 1/2 *g *(2/Cosθ ) = g/Cosθ
Solving gives Sinθ Cosθ = Sin(2*θ) = g/490 = 9.8/490 = .02
(Note the identity Sinθ Cosθ = Sin(2*θ))

sin-1(.02) = 1.15 degrees

But that = 2θ,
so I'd say θ = (1.15)/2 degrees = .575 degrees
 
  • #3
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Vy*t = 1/2 g*t^2 can't be right though, because that leaves out the initial y velocity of the bullet, right? I'm thinking it would have to be Vy*t = 1/2 g*t^2 + V(0)sin(theta)
 
  • #4
LowlyPion
Homework Helper
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Vy*t = 1/2 g*t^2 can't be right though, because that leaves out the initial y velocity of the bullet, right? I'm thinking it would have to be Vy*t = 1/2 g*t^2 + V(0)sin(theta)

The V*Sinθ term is the Vy term
 
  • #5
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Oh, ok, I get it now. Thank you very much!
 

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