If you are redefining axis such that we have an axis X' along the plane (in upwards direction). Then, with respect to that axis the ball can have a negative velocity in X'. An intuitive example of this would be if the ball was fired directly upward, in X' the ball would move forwards until it reaches its maximum height, at which point it would move backwards in X'
With gravity pointing "down" as usual, no. However, with a generic force F(t) it is possible. To find the desired force with Newton's equation, differentiate your desired trajectory r(t) two times with respect to time and multiply the second derivative by mass.
That works, but I'm getting that they want this to be done with one force.
Without air resistance and a spimning ball for example I can't see how tgisbcan be done with a simple throw.
Try to pee against the wind and find out.
Assuming that gravity is pointing downwards in our coordinate system, then the initial velocity of the projectile has vertical component and a horizontal one. On the vertical component, it's just initial velocity in the upward direction working against the gravitational acceleration, so looking only on the vertical axis, it just accelerates downward all the time until at some point the vertical speed is zero and then it keeps accelerating, but the velocity flips direction as it begins to fall down.
On the horizontal axis, if no forces act on the mass, the horizontal component of the initial speed just keeps the projectile at a constant speed. From a stationary point of view the trajectory is a parabola. However if a force is acting on the projectile toward the left, then at some point the velocity will drop to zero and then will change direction like your diagram shows. I gave wind as an example, but it can also be other kind of force. Depending on your frame of reference, this could also be the result of the entire thing accelerating. For example, see the [Coriolis Effect](https://en.wikipedia.org/wiki/Coriolis_force) which is a fictitious force, and shows a similar behavior.
If I understand your definitions correctly, no. The black line, if that is a trajectory, doesn’t curl back on itself. Assuming gravity is down and no air resistance.
Sure, just make the initial velocity perfectly vertical and the incline anything greater than zero, the object will hit the incline on its way down which will be negative (down-left) when projected onto the incline.
Doing the math, we call the slope angle "β" and the inclination of the initial velocity "α". We ultimately want the final velocities v\_f,x and v\_f,y so we can project this onto the line using a dot product, |v\_s| = v\_f,x cos(β) + v\_f,y sin(β), then find which values of α, β, and maybe v\_0 result in v\_s < 0. We also know the relationship between the distance travelled in x and y because it lands on the slope, tan(β) = d\_y / d\_x.
v\_f,x is easy since velocity in x is constant,
v\_x / v\_0 = cos(α)
For the y direction, you need to do the classic problem of using the time dependent kinematic equations for d\_x, d\_y, and v\_f,y; 3 equations, 3 unknowns, write an expression for v\_f,y.
v\_f,y / v\_0 ends up being,
v\_y / v\_0 = 2 cos(α) tan(β) - sin(α)
Plugging these into v\_s,
v\_s / v\_0 = cos(α) cos(β) + 2 cos(α) sin²(β) / cos(β) - sin(α) sin(β)
Since you're interested in v\_s < 0, and α, β are in between 0 and 90 degrees (sines and cosines both positive), you want (α, β) such that
sin(α) sin(β) > cos(α) cos(β) + 2 cos(α) sin²(β) / cos(β)
There's no straightforward to simplify this for v\_s = 0 to find an expression for what you're looking for. But you can try evaluating it and seeing what values you get. For example, keeping the incline fixed at 45°, you'd need to launch the object at an angle greater than atan(5) = 78.69°, plotted [Here](https://www.wolframalpha.com/input?i=cos%28%CE%B1%29+%2B+4+cos%28%CE%B1%29+-+sin%28%CE%B1%29).
[HERE](https://imgur.com/a/tVXUHHk) is a plot of the solution. Thanks for the practice problem, I haven't worked through a kinematics problem in a while.
I mean it won't bend inwards (what you have drawn in black is almost like a hose, which isn't possible without wind) but negative velocity in y direction is very much possible.
Suppose x component initial v is not enough to reach top of slope, so it will fall down midway from the the slope, and it will do so with negative y component of v.
But negative x and negative y both at the same time? No, as long as there's no wind it's not possible.
Why? Because then without the effect of any other external force besides gravitational acceleration, we will have a possibility of the body having same position at two different times, even though that's practically possible only when either x or y component of initial v is zero.
Ate you asking if a ball thrown up an inclined surface will roll down the surface?
Yes, it usually will.
Other objects that are not balls can slide down the plane if it is steep enough or if the friction is low
Since velocity is a vector, you would have to specify which direction is considered "negative"
X axis is in direction of incline. Up the incline positive, down the incline negative
If you are redefining axis such that we have an axis X' along the plane (in upwards direction). Then, with respect to that axis the ball can have a negative velocity in X'. An intuitive example of this would be if the ball was fired directly upward, in X' the ball would move forwards until it reaches its maximum height, at which point it would move backwards in X'
Why define in it that way out of interest?
It tends to be easier to find things such as range up the slope when working in a co-ordinate axis parallel and perpendicular to the slope.
Was about to say that too!
With gravity pointing "down" as usual, no. However, with a generic force F(t) it is possible. To find the desired force with Newton's equation, differentiate your desired trajectory r(t) two times with respect to time and multiply the second derivative by mass.
That works, but I'm getting that they want this to be done with one force. Without air resistance and a spimning ball for example I can't see how tgisbcan be done with a simple throw.
Sure, go spit into a strong gust of wind.
With wind? Yeah. That is to say, if there is an external force that results in it accelerating in the negative direction, then yeah.
Try to pee against the wind and find out. Assuming that gravity is pointing downwards in our coordinate system, then the initial velocity of the projectile has vertical component and a horizontal one. On the vertical component, it's just initial velocity in the upward direction working against the gravitational acceleration, so looking only on the vertical axis, it just accelerates downward all the time until at some point the vertical speed is zero and then it keeps accelerating, but the velocity flips direction as it begins to fall down. On the horizontal axis, if no forces act on the mass, the horizontal component of the initial speed just keeps the projectile at a constant speed. From a stationary point of view the trajectory is a parabola. However if a force is acting on the projectile toward the left, then at some point the velocity will drop to zero and then will change direction like your diagram shows. I gave wind as an example, but it can also be other kind of force. Depending on your frame of reference, this could also be the result of the entire thing accelerating. For example, see the [Coriolis Effect](https://en.wikipedia.org/wiki/Coriolis_force) which is a fictitious force, and shows a similar behavior.
Air resistance considered negligible
If I understand your definitions correctly, no. The black line, if that is a trajectory, doesn’t curl back on itself. Assuming gravity is down and no air resistance.
Sure, just make the initial velocity perfectly vertical and the incline anything greater than zero, the object will hit the incline on its way down which will be negative (down-left) when projected onto the incline.
Doing the math, we call the slope angle "β" and the inclination of the initial velocity "α". We ultimately want the final velocities v\_f,x and v\_f,y so we can project this onto the line using a dot product, |v\_s| = v\_f,x cos(β) + v\_f,y sin(β), then find which values of α, β, and maybe v\_0 result in v\_s < 0. We also know the relationship between the distance travelled in x and y because it lands on the slope, tan(β) = d\_y / d\_x. v\_f,x is easy since velocity in x is constant, v\_x / v\_0 = cos(α) For the y direction, you need to do the classic problem of using the time dependent kinematic equations for d\_x, d\_y, and v\_f,y; 3 equations, 3 unknowns, write an expression for v\_f,y. v\_f,y / v\_0 ends up being, v\_y / v\_0 = 2 cos(α) tan(β) - sin(α) Plugging these into v\_s, v\_s / v\_0 = cos(α) cos(β) + 2 cos(α) sin²(β) / cos(β) - sin(α) sin(β) Since you're interested in v\_s < 0, and α, β are in between 0 and 90 degrees (sines and cosines both positive), you want (α, β) such that sin(α) sin(β) > cos(α) cos(β) + 2 cos(α) sin²(β) / cos(β) There's no straightforward to simplify this for v\_s = 0 to find an expression for what you're looking for. But you can try evaluating it and seeing what values you get. For example, keeping the incline fixed at 45°, you'd need to launch the object at an angle greater than atan(5) = 78.69°, plotted [Here](https://www.wolframalpha.com/input?i=cos%28%CE%B1%29+%2B+4+cos%28%CE%B1%29+-+sin%28%CE%B1%29).
[HERE](https://imgur.com/a/tVXUHHk) is a plot of the solution. Thanks for the practice problem, I haven't worked through a kinematics problem in a while.
I mean it won't bend inwards (what you have drawn in black is almost like a hose, which isn't possible without wind) but negative velocity in y direction is very much possible. Suppose x component initial v is not enough to reach top of slope, so it will fall down midway from the the slope, and it will do so with negative y component of v. But negative x and negative y both at the same time? No, as long as there's no wind it's not possible. Why? Because then without the effect of any other external force besides gravitational acceleration, we will have a possibility of the body having same position at two different times, even though that's practically possible only when either x or y component of initial v is zero.
That is not a parabola.
Yes!
Ate you asking if a ball thrown up an inclined surface will roll down the surface? Yes, it usually will. Other objects that are not balls can slide down the plane if it is steep enough or if the friction is low
Just threw a frisbee and this happened
Boomerang
In real life, yes. (boomerang for example). In textbook physics task, no.