# Conceptual question regarding velocity in projectile motion vs. up a ramp

I have a question regarding conservation of energy in regards to two different scenarios:

In the first case I was told that a ball with an initial velocity, V, was launched at an angle, $$\theta$$ and to solve for the max height in terms of $$h$$.

So, using $$KE_i + U_i = KE_f + U_f$$

I found:

$$KE_i= \frac 12 mv^2$$

$$U_i = 0$$

$$KE_f= \frac 12 m(vcos\theta)^2$$

$$U_f = mgh$$

so $$h$$ was $$\frac {[v^2(1 - cos^2\theta)]}{2g}$$

Now...

the next question had the same parameters, except it was up an incline (frictionless). I got,

$$KE_i = \frac 12 mv^2$$

$$U_i = 0$$

$$KE_f = 0$$ (why?)

$$U_f = mgh$$

and they explained it with:

"Interestingly, the answer does not depend on $$\theta$$. The difference between this situation and the projectile case is that the ball moving up a slope has no kinetic energy at the top of its trajectory whereas the projectile launched at an angle does."

But why? why does the final kinetic energy of the projectile motion depend on the angle but the final kinetic energy of a ball being launched up a slope is 0, and doesn't depend on the angle?

• The ball comes to rest on the incline...? The projectile does not come to rest at the top of its trajectory. Mar 9 '19 at 6:56

From a comment:

I'm curious as to why the final kinetic energy of the projectile motion depends on the angle but the final kinetic energy of a ball being launched up a slope is 0, and doesn't depend on the angle (at max height for both)

I think this comment shows a misunderstanding that need to be addressed.

There word "final" is subjective and is up for us to define. For energy conservation problems, you pick instants in time you want to compare, and then you look at the energies at each instant, and then it can say the total energy is equal for each instant. Therefore, it makes no sense to compare the "final" energy of one system to the "final" energy of a completely different system. You have to take what you know about each system first, and then you look at the energy.

For example, with the projectile we subjectivity pick "initial" to be right at launch and "final" to be when the projectile is at its maximum height. Then we say "ok. I know that at the initial point the energy is all kinetic. I know at the maximum height there is potential energy and some kinetic energy given by what I know the velocity to be. I know these total energies must be equal."

With the incline we subjectivity pick "initial" to be right at launch and "final" to be when the projectile is at its maximum height. Then we say "ok. I know that at the initial point the energy is all kinetic. I know at the maximum height there is potential energy and no kinetic energy given by what I know the velocity to be. I know these total energies must be equal."

There isn't any reason to think that these systems should behave the same "at maximum height" just because we subjectively choose the "final point" to be at maximum height.

The difference is that the motion of the projectile is constrained by one force - its weight acting vertically downwards and there is no horizontal forc,e whereas the object moving up the ramp is constrained by two forces - its weight acting vertically downwards and the normal reaction force due to the ramp which has a horizontal component.

The normal reaction force does no work as it is at right angles to the motion of the object but it does mean that at all times $$v_{\rm ramp}^2 = v_{\rm ramp,vertical}^2+v_{\rm ramp,horizontal}^2 = v_{\rm ramp}^2 \sin^2 \alpha + v_{\rm ramp}^2 \cos^2 \alpha$$ where $$\alpha$$ is the angle of the ramp to the horizontal.
The horizontal and vertical motions are not independent of one another and the direction of the velocity is always at an angle $$\alpha$$ to the horizontal.

For the projectile the horizontal component of its velocity does not change but its vertical component of velocity does - they are independent of one another.

So when the direction of the velocity of the projectile is at an angle $$\beta$$ to the horizontal $$v_{\rm projectile}^2 = v_{\rm projectile,vertical}^2+v_{\rm projectile,horizontal}^2 = v_{\rm projectile}^2 \sin^2 \beta + v_{\rm projectile}^2 \cos^2 \beta$$ with $$v_{\rm projectile} \cos \beta = v_{\rm projectile,initial} \cos \theta = \rm constant$$, where $$\theta$$ was the initial direction of projection to the horizontal.
The direction of the velocity to the horizontal changes with time and at the greatest height there is still a horizontal component of velocity.

Your question wasn't clear, for Q1, there's a "common sense" of where to launch for the highest point, especially when they were teaching energy conservation. For Q2, I think you were suppose to use standard Newtonian and simply integrate it out.

• I'm curious as to why the final kinetic energy of the projectile motion depends on the angle but the final kinetic energy of a ball being launched up a slope is 0, and doesn't depend on the angle (at max height for both) Mar 9 '19 at 6:51
• @Jake3017 at maximum high of a single $\theta$, there might be a kinematic component, at maximum height of all $\theta$, there's kinematic energy =0 as well.
– J C
Mar 9 '19 at 17:59

Gravitational potential energy is a function of solely vertical motion of an object, so in a case where a body has both vertical and horizontal component in its velocity vector (at an angle), we only deal with the vertical component when calculating the kinetic energy that is being transformed to potential energy at the peak height. Thus the angle doesn't matter.

In your first case, the vertical velocity at maximum height is zero, so making $$KE_f= \frac 12 m(vcos\theta)^2$$ and not zero when dealing with vertical partition of the motion is where your problem is emanating from. Based on the law of conservation of energy, the kinetic energy computed with the vertical component of the obiect's velocity vector from the start of the motion is transformed into potential energy at the peak height of the flight/trajectory (the velocity remaining in the body is the horizontal component), so at apogee the object stops moving vertically and vertical velocity is zero so kinetic energy for the vertical component is zero. In the inclined ramp case, the body stops moving both horizontally and vertically at the peak height (as long as it doesn't fall off the top of the ramp and follow a parabolic trajectory down, the kinetic energy at the apogee in this case is zero (both vertical and horizontal)), but we still use the vertical component of the velocity to calculate the kinetic energy being transformed to potential regardless.

• The work done by the OP is correct Mar 9 '19 at 17:50
• @ Stevens. My answer didn't go against that, but since the OP is in search of potential energy at the max height, the horizontal energy of the body shouldn't come in anymore. Mar 9 '19 at 18:13
• Your very first sentence says they broke conservation of energy. Energy isn't a vector. There isn't horizonal and vertical energy Mar 9 '19 at 20:30
• I never said there was, but if they're not distinguished or categorized, things become complicated. You can see the OP runs into problem with the first case because his final kinetic energy is derived from the horizontal velocity of the object while the vertical velocity is zero. This definitely is something that needs to be contrasted so OP will understand which velocity is being transformed to potential at the apogee. Mar 9 '19 at 21:08
• I guess you are just using words differently than I am used to. You say things like "vertical kinetic energy", and you seem to say that the OP is wrong in saying that the final kinetic energy is $\frac12m(v\cos\theta)^2$ when this is in fact the kinetic energy of the projectile at maximum height. There is no energy conservation broken here. Mar 9 '19 at 21:29