Finding Projectile Motion range at an angle?

In summary, you found the range of a toy cannon when it is placed on a ramp on a hill, and you can find the distance along the hill by relating the x-value you found with the the the term [g] which is the slope of the hill.
  • #1
Kavorka
95
0
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!
 
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  • #2
You are looking for an intersection of a straight line with a parabola. Do you know howto find it?
 
  • #3
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
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.
 
  • #5


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
 

FAQ: Finding Projectile Motion range at an angle?

What is projectile motion?

Projectile motion is the motion of an object through the air that is affected by gravity. It follows a curved path instead of a straight line due to the force of gravity acting on it.

How do you find the range of a projectile at a given angle?

To find the range of a projectile at a given angle, you can use the formula: R = (v^2 * sin(2θ)) / g, where R is the range, v is the initial velocity, θ is the angle, and g is the acceleration due to gravity. Alternatively, you can also use a projectile motion calculator or plot the trajectory of the projectile on a graph to determine the range.

What is the optimal angle for maximum range in projectile motion?

The optimal angle for maximum range in projectile motion is 45 degrees. This is because at this angle, the vertical and horizontal components of the initial velocity are equal, resulting in the longest range for the same initial velocity.

How does air resistance affect projectile motion range?

Air resistance can decrease the range of a projectile by slowing it down during its flight. This is because air resistance creates an opposing force that acts in the opposite direction of the projectile's motion, reducing its horizontal velocity and therefore its range.

Can the range of a projectile be negative?

Yes, the range of a projectile can be negative if it is launched at an angle greater than 90 degrees. In this case, the projectile will travel downwards and the range will be below the horizontal starting point. However, in most cases, the range of a projectile will be positive as it is launched at an angle between 0 and 90 degrees.

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