Finding Projectile Motion range at an angle?

[Kavorka]

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!

 

[voko]

You are looking for an intersection of a straight line with a parabola. Do you know howto find it?

 

[Kavorka]

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.

 

[voko]

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.

 

[HalfLostAndSearching]

I would first clarify the problem by drawing a diagram to better understand the situation. From the given information, it seems that the toy cannon is placed on a ramp with a slope of angle B, while the hill itself has a slope of angle A. The cannonball is launched with a muzzle speed of v and the goal is to find the range R of the cannonball along the hill, not along the x-axis.

To solve this problem, we can use the equation for projectile motion on a flat surface, R = (v^2/g)sin(2θ), where θ is the angle of launch with respect to the horizontal. However, in this case, we need to take into account the angles of both the ramp and the hill.

First, we need to find the angle of launch, θ. From the diagram, we can see that the angle of launch is equal to the angle of the ramp, B. This is because the cannonball is launched along the ramp, not along the hill. So, we can rewrite the equation as R = (v^2/g)sin(2B).

Next, we need to consider the effect of the hill's slope, A, on the range. We can do this by breaking down the range into two components - one along the ramp and one along the hill. The component along the ramp is given by R_ramp = R*cos(A), and the component along the hill is given by R_hill = R*sin(A). Therefore, we can rewrite the equation as R = R_ramp + R_hill = (v^2/g)sin(2B)cos(A) + (v^2/g)sin(2B)sin(A).

We can simplify this equation by using the double angle formula for sine, sin(2x) = 2sin(x)cos(x). This gives us R = (2v^2/g)sin(B)cos(B)cos(A) + (2v^2/g)sin(B)cos(B)sin(A).

Now, we can use the trigonometric identity, cos(A-B) = cos(A)cos(B) + sin(A)sin(B), to further simplify the equation. This gives us R = (2v^2/g)sin(B)cos(A-B). Finally, we can use the trigonometric identity, sin(x) = tan(x)/√(1+tan^2