Finding Projectile Motion range at an angle?

  • Thread starter Kavorka
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  • #1
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The original problem has very confusing wording without the picture so I reworded it for simplicity:

A toy cannon is placed on a ramp on a hill, pointing up the hill. With respect to the x-axis the hill has a slope of angle A and the ramp has a slope of angle B. If the cannonball has a muzzle speed of v, show that the range R of the cannonball (as measured up the hill, not along the x-axis) is given by:

R = [2v^2 (cos^2 (B))(tan(B) - tan(A))] / [g cos(A)]

The base equation we've derived for projectile range on a flat surface:
R = (v^2 /g)sin(2θ­)
from the parabolic equation:
y = vt + (1/2)at^2 (where v is initial velocity, and v and a are in the y-direction)
and setting y to 0.

I'm not completely sure how to correctly start this problem or how to properly take the angle of the slope of the hill into account, the trig is a bit overwhelming. Even a good shove in the right direction would help immensely!
 

Answers and Replies

  • #2
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You are looking for an intersection of a straight line with a parabola. Do you know howto find it?
 
  • #3
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Yes, but were not analyzing it graphically we're analyzing it with motion equations and trig. Since I posted I was able to find an equation for time by finding the x and y-values of where the cannonball lands in terms of motion equations and relating them with the tangent of the slope of the hill, and solving for time. I then plugged T into Range = initial x velocity * time and combined the terms. My answer comes out to exactly what the original equation I want to derive is except it is over the term [g] not [g cos(A)]. I'm not sure where that cosine of the slope of the hill comes from.
 
  • #4
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391
You found the value of ##x## where the ball touches the hill. Well done! But you are asked to find the distance along the hill. It is very simply related with the the x-value you found. You are just a step away from the correct answer.
 

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