Urgent: Angle of Projectile

  • Thread starter ericka141
  • Start date
  • #1
8
0

Homework Statement


You shoot a cannon at a distant cliff. The cliff is 170m high and 550m away, with the launch speed of the projectile being 110 m/s. Wind resistance is ignored. At what angles, relative to the horizontal, should the projectile be shot to hit the target?

Hint: 1/cos2θ = 1+tan2θ


Homework Equations


Kinematics Equations:
vf=vi + a*t
d= vi*t + 1/2a*t^2
d= (vf^2-vi^2)/2a

The Attempt at a Solution


I've broken the problem up into x and y components:
vix= vixcos(θ)
dx= 550m (or vixcos(θ)*t)
ax= 0m/s^2

viy= viysin(θ)
dy= 170m (or viysin(θ)*t)
ay= -9.81m/s^2

I know how to find the maximum distance that a projectile travels when given the initial velocity (or the distance when given the initial velocity and the angle it's shot at) but I'm not sure how to find these angles. I'm not exactly sure where tan(θ) comes in. If somebody could help me with this that would be GREAT, this assignment is due tonight.
 

Answers and Replies

  • #2
56
2
If you know x and y (or [itex]v_x[/itex] and [itex]v_y[/itex]), you can find the angle. Who is the man? Tan is the Man!

Recall SOH CAH TOA
 
  • #3
8
0
Then how would I use 1/cos2θ=1+tan2θ ??
 
  • #4
56
2
You don't. [itex]tan(\theta) = \frac{d_x}{d_y}[/itex]

Solve for [itex]\theta[/itex]
 

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