# Projectile motion on an inclined plane

• danyull
In summary, to hit the inclined plane at an angle of θ above the horizontal, the ball should be launched with an initial speed of v0 and an angle of π/6 above the inclined plane.
danyull

## Homework Statement

A ball is thrown with initial speed v0 up an inclined plane. The inclined plane makes an angle of π/6 above the horizontal line and the ball is launched at an angle θ above the inclined plane. No air resistance in this problem.
(a) How long does the ball stay in the air?
(b) At what angle θ should the ball be launched in order to fall back on the plane normal to the inclined surface?

## Homework Equations

Newton's laws and kinematics equations?

## The Attempt at a Solution

I tried making a rotated reference frame (expressed as a matrix transformation below) centered at the launch point where $$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \cos(π/6) & \sin(π/6) \\ -\sin(π/6) & \cos(π/6) \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \\ \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \frac {\sqrt{3}} 2 & \frac 1 2 \\ - \frac 1 2 & \frac {\sqrt{3}} 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$$
and since the force of gravity is ##\vec {\mathbf F} = -mg ~ \hat y##, then in the new reference frame $$F_x' = \frac {\sqrt{3}} 2 F_x + \frac 1 2 F_y = - \frac 1 2 mg \\ F_y' = - \frac 1 2 F_x + \frac {\sqrt{3}} 2 F_y = - \frac {\sqrt{3}} 2 mg$$
and acceleration ##a_x'## and ##a_y'## will just be the above two forces with mass divided out.

From here on I'm a bit stuck. As for part (a), I want to use the kinematic equations with my given initial velocity. At first I thought to solve for the time it takes to reach the peak of its trajectory and then multiplying by 2, but I realized that would be the time it takes to hit the ground if the inclined plane weren't there. My next guess was solving for the time when the horizontal velocity reached zero since in this reference frame there's a negative horizontal acceleration, but I'm not sure that would be the exact time that the ball hits the inclined plane.

Any help would be appreciated!

danyull said:

## The Attempt at a Solution

I tried making a rotated reference frame (expressed as a matrix transformation below) centered at the launch point where $$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \cos(π/6) & \sin(π/6) \\ -\sin(π/6) & \cos(π/6) \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \\ \begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} \frac {\sqrt{3}} 2 & \frac 1 2 \\ - \frac 1 2 & \frac {\sqrt{3}} 2 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$$
and since the force of gravity is ##\vec {\mathbf F} = -mg ~ \hat y##, then in the new reference frame $$F_x' = \frac {\sqrt{3}} 2 F_x + \frac 1 2 F_y = - \frac 1 2 mg \\ F_y' = - \frac 1 2 F_x + \frac {\sqrt{3}} 2 F_y = - \frac {\sqrt{3}} 2 mg$$
and acceleration ##a_x'## and ##a_y'## will just be the above two forces with mass divided out.
OK. That's a very formal way to do it, but it works! You could also just project the acceleration due to gravity along your x' and y' axes.

From here on I'm a bit stuck. I want to use the kinematic equations with my given initial velocity. At first I thought to solve for the time it takes to reach the peak of its trajectory and then multiplying by 2, but I realized that would be the time it takes to hit the ground if the inclined plane weren't there.
Right.

My next guess was solving for the time when the horizontal velocity reached zero since in this reference frame there's a negative horizontal acceleration, but I'm not sure that would be the exact time that the ball hits the inclined plane.
Your idea that the x'-component of velocity should be zero at impact might be right for part (b). But your reason for stating that doesn't make sense to me. Keep in mind that for part (b) you want the ball to hit the incline perpendicularly to the incline.

For part (a), think about the coordinates of the point of impact in your primed coordinate system.

danyull
Thanks for your input! My professor wants us to practice rotated reference frames using matrix transformations, hence this homework problem.

I think I figured out part (a). I just used the equation ##y' = v_{0,y}' t + \frac 1 2 a_y' t^2## and set it equal to 0, then solve for ##t##. I'm still working out part (b).

Last edited:
danyull said:
I think I figured out part (a). I just used the equation ##y' = v_{0,y}' t + \frac 1 2 a_y' t## and set it equal to 0, then solve for ##t##. I'm still working out part (b).
Good. (Forgot to type the square on the t in the last term?)

danyull
Ah, right I did. Thank you!

I believe I have the second part too now. I'll just set ##\vec {\mathbf r}' \cdot \hat {\mathbf y} = 0## because the dot product is zero when the vectors are orthogonal. The x-component drops out, and the position ##y'## will be from the equation I just used, evaluated at the time that I solved for.

danyull said:
I believe I have the second part too now. I'll just set ##\vec {\mathbf r}' \cdot \hat {\mathbf y} = 0## because the dot product is zero when the vectors are orthogonal.
So, you are saying that the position vector is perpendicular to the y'-axis at impact? (I'm interpreting ##\vec {\mathbf r}'## to be the position vector in the primed frame with origin at the launch point.) Wouldn't that always be true for any impact location on the plane (even if the ball does not hit perpendicularly to the plane)?

The x-component drops out, and the position ##y'## will be from the equation I just used, evaluated at the time that I solved for.
But you already know that ##y'## will be zero at impact.

Try to formulate the idea that the ball is moving perpendicular to the inclined surface at impact.

danyull
Oops, yeah I realized that wouldn't work after trying it. Do you think I could set ##\vec {\mathbf v}' \cdot \hat {\mathbf x}' = 0## instead? Evaluating the velocity at the time I solved for, the vector points "down" in the coordinate system, which is perpendicular to the coordinate's unit vector x. Then I solve for θ.

danyull said:
Oops, yeah I realized that wouldn't work after trying it. Do you think I could set ##\vec {\mathbf v}' \cdot \hat {\mathbf x}' = 0## instead? Evaluating the velocity at the time I solved for, the vector points "down" in the coordinate system, which is perpendicular to the coordinate's unit vector x. Then I solve for θ.
Yes, sounds like a good plan.

danyull

## 1. What is projectile motion on an inclined plane?

Projectile motion on an inclined plane is the motion of an object that is launched at an angle on a surface that is not horizontal. The object will follow a curved path due to the influence of gravity and the incline of the plane.

## 2. What factors affect projectile motion on an inclined plane?

The factors that affect projectile motion on an inclined plane include the angle of the incline, the initial velocity of the object, and the force of gravity. The mass and shape of the object may also have an impact on its motion.

## 3. How do you calculate the range of a projectile on an inclined plane?

The range of a projectile on an inclined plane can be calculated using the equation R = (v^2 * sin2θ) / g, where R is the range, v is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

## 4. How does the angle of the incline affect the motion of a projectile?

The angle of the incline can affect the motion of a projectile in several ways. A steeper incline will result in a shorter range and a higher trajectory, while a shallower incline will result in a longer range and a lower trajectory. The angle can also affect the time of flight and the maximum height reached by the projectile.

## 5. What is the difference between projectile motion on an inclined plane and on a horizontal surface?

The main difference between projectile motion on an inclined plane and on a horizontal surface is the presence of a gravitational force acting on the object. On an inclined plane, the gravitational force will act in a downward direction along the incline, causing the object to follow a curved path. On a horizontal surface, there is no incline and the gravitational force will act straight down, resulting in a simpler, straight-line motion.

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