# Horizontal Range (with incline)

• jesuslovesu
In summary, the ball has a maximum possible range up an incline if its launch angle is phi and the angle of the incline is theta.
jesuslovesu
[SOLVED] Horizontal Range (with incline)

## Homework Statement

A ball is thrown up an (v0) incline. The ball is thrown at an angle phi from the incline and the incline has an angle theta with the ground.
1) Show the ball lands a distance $$R = 2{v_0}^2 sin\theta cos(\theta + \phi ) / (gcos{\phi }^2$$

2) Show that for a given v0 and phi, the maximum possible range up the incline is $$R_{max} = {v_0}^2/(g(1+sin\phi))$$

## The Attempt at a Solution

I was able to do 1) fairly easily, however, I'm having a bit of trouble with 2).
I was thinking that since the max range on a horizontal plane is 45 degrees, I could just shift theta by 45 degrees and say $$\theta = 45 + \phi$$ unfortunately, that doesn't seem to lead me to the correct answer.

I was then thinking of taking the derivative of R and seeing if I could find the maximum of it that way.

$$dR/d\theta = cos\theta cos(\theta + \phi) - sin \theta sin( \theta + \phi ) = 0$$
Unfortunately this seems to lead me nowhere (don't know how to solve for theta)

Does anyone know how I should go about finding theta? I was hoping that it was going to be 45 degrees, but I don't think that it is, and now I don't know where to go.

Hi jesuslovesu,

Could the statement of the problem be wrong here? It states that the angle phi is the launch angle of the ball relative to the incline, and theta is the angle of the incline itself (relative to the ground). Were those supposed to be switched?

If those definitions were true, then for level ground, theta=0, which means that your range formula in part 1 gives zero (since it has a sin(theta) in the numerator).

Oh my bad, sorry yes, theta is the angle of the ball's velocity and phi is the angle of the incline.

$$s=Rcos\phi$$
$$v_{YO}=v_{0}sin(\phi+\theta)$$
$$v_{XO}=v_{0}cos(\phi+\theta)$$
$$h=v_{YO}t-\frac{gt^{2}}{2}$$
$$s=v_{XO}t$$
so
$$t=\frac{Rcos\phi}{v_{0}cos(\phi+\theta)}$$
so
$$\frac{gR^{2}cos^{2}\phi}{2v_{0}^{2}cos^{2}(\phi+\theta)}+Rsin\phi-\frac{Rsin(\phi+\theta)cos\phi}{cos(\phi+\theta)}=0$$
so
$$R=0$$ or $$R=\frac{2v_{0}^{2}cos(\phi+\theta)sin\theta}{cos^{2}\phi}$$

$$\frac{dR}{d\theta}=cos\theta{cos(\phi+\theta)}-sin\theta{sin(\phi+\theta)}=0$$
but
$$cos(\phi+\theta)=cos\theta{cos\phi}-sin\theta{sin\phi}$$
$$sin(\phi+\theta)=sin\phi{cos\theta}+cos\phi{sin\theta}$$
to equation
$$\frac{dR}{d\theta}=cos\phi(cos^{2}\theta-sin^{2}\theta)-2sin\theta{cos\theta}sin\phi=0$$
and
$$cos2\theta=cos^{2}\theta-sin^{2}\theta$$
$$sin2\theta=2sin\theta{cos\theta}$$
so
$$tg2\theta=ctg\phi$$

Last edited:

## 1. What is the definition of horizontal range?

Horizontal range is the maximum distance a projectile travels horizontally from its initial position to the point where it hits the ground.

## 2. How is horizontal range affected by incline?

Incline can affect horizontal range by changing the initial velocity and angle of the projectile, which can result in a longer or shorter horizontal distance traveled.

## 3. What is the formula for calculating horizontal range with incline?

The formula for calculating horizontal range with incline is: R = (v^2 * sin2θ)/g, where R is the horizontal range, v is the initial velocity, θ is the angle of incline, and g is the acceleration due to gravity.

## 4. Can horizontal range be negative?

No, horizontal range cannot be negative. It is always measured as a positive distance from the initial position to the point of impact.

## 5. How does air resistance impact horizontal range?

Air resistance can decrease the horizontal range of a projectile by slowing it down as it travels through the air. This can be accounted for in the calculation of horizontal range by adjusting the initial velocity term.

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