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Two-Dimensional Projectile Motion

  • Thread starter Haikon
  • Start date
  • #1
2
0

Homework Statement



An artillery shell is fired at mach 2 at an enemy encampment located at a range of 1700m. The initial position of the shell is 72 meters below the encampment, and is aimed at an angle θ.
Solve the problem for θ.

We can get the following variables/data from this:

Range = 1700m
Vertical displacement = 72m
Initial speed = 800m/s

Homework Equations



From this I derived these equations:

1700=800cosθt (d = r*t)
72 = 800sinθt - 4.905t^2 (d = Vi*t + 1/2a*t^2)

2 equations with two unknowns. Problem is, I don't know how to solve them.

The Attempt at a Solution



What I attempted was solving for t in the first equation, giving me t = (1700)/(800cosθ). Plugging this back into the second equation, I get 72 = 800sinθ(1700/800cosθ) - 4.905(1700/800cosθ)^2

From here, I got crazy solutions that ended up putting me at 89.5772 degrees for θ, which is incorrect. How can I correctly solve this?
 

Answers and Replies

  • #2
1,065
10
Taking upward as positive, your vertical displacement(below) should be negative.
 
  • #3
2
0
No, the shell is traveling up and landing 72 meters above its original position, so vertical displacement will be +72. Both range and vertical displacement are positive in this case. Here is the diagram (drawn in paint and not to scale):

http://puu.sh/19JvY [Broken]
 
Last edited by a moderator:
  • #4
1,065
10
Sorry for my interpretation.
To solve the equation, you use the identity, 1/Cos2θ=1+tan2θ
There must be 2 roots, since they are the points on upward and downward path.
 
Last edited:

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