# Speed of the particle at a height h

1. Sep 15, 2016

### moenste

1. The problem statement, all variables and given/known data
A particle is projected with speed v at an angle α to the horizontal. Find the speed of the particle when it is at a height h.

2. The attempt at a solution
a. Since PE gained is KE lost we need to find KE and PE.
KE = 0.5 * m * (v sin α)2 (the vertical component of velocity)
PE = mgh

0.5m * (v sin α)2 = mgh
v = √ 2gh / (sin α)2

or

b. v2 = u2 + 2as
Upward direction is positive. So we have:
u = v sin α
a = -g
s = h
v = √(v sin α)2 -2gh

But both of them do not fit the answer.

Thanks in advance for any help.

Last edited: Sep 15, 2016
2. Sep 15, 2016

### Staff: Mentor

A particle's speed is tied to its total KE. You don't have to know how this energy is apportioned to horizontal and vertical components, simply the total energy. However, you do know how the KE changes with a change in height, h. Can you write an expression for the initial KE at height zero? How about for the total KE at some height h?

3. Sep 15, 2016

### moenste

KE at zero height = 0.5 mv2
KE at some height is zero and equals to PE = mght?

4. Sep 15, 2016

### Staff: Mentor

Yes, the initial KE is specified by its initial velocity: $KE_o = \frac{1}{2}m v_o^2$ .

But the change in KE depends upon the change in height. Can you write an expression for KE(h)? Assume that KE=KEo when h=0.

5. Sep 15, 2016

### moenste

Update: I worked on the b. part and this is what I got:

Consider the vertical motion: v2 = u2 + 2as
v2 = (v sin α)2 - 2gh

Consider horizontal motion: v = v cos α

Since we need to find the speed, by Pythagoras: speed2 = vvertical2 + vhorizontal2
So speed = √(v sin α)2 - 2gh + (v cos α)2 = √v2(sin2 α + cos2 α) - 2gh = √v2 - 2gh

Does this look correct?

6. Sep 15, 2016

### moenste

Hm, if we assume that a particle does not reach its maximum height (at max height the velocity is zero) and we use the fact that KE lost = PE gained, I would write the formula like this: KE0-KE = PE. So KE0 = when the particle is on the ground less the KE when the particle is somewhere flying equals to PE. So 9 - 3 = 6. 9 was the initial KE0, 3 is the KE at some height and 6 is the KE lost or PE gained. That could be written like 0.5 mvinitial2 - 0.5mvnew2 = mghat this moment. Am I following you correctly?

Update: I think I got it what are you suggesting :).

I need vnew. So, I put - 0.5mvnew2 = mghat this moment - 0.5 mvinitial2
We simplify it:
0.5vnew2 = 0.5vinitial2 - gh
vnew = √vinitial2 - 2gh

7. Sep 15, 2016

### Staff: Mentor

That's the idea. The KE at any height h (where h is bounded by the trajectory of the particle) is given by $KE_h = KE_o + \Delta KE$.

You know what the initial KE is: $KE_o = \frac{1}{2}m v^2$; and you know how $\Delta KE$ is tied to $\Delta PE$. Combine the pieces and write it out. Then the speed follows from the "translation" of KE to speed in the usual manner.

8. Sep 15, 2016

### Staff: Mentor

That works, but it's certainly going about it the hard way. You don't have to separate the velocity components when you are dealing with the change in energy.

9. Sep 15, 2016

### moenste

Though it's better for the understanding. As soon as I wrote it out I understood right away the logic behind it. Thank you.

10. Sep 15, 2016

### Staff: Mentor

That's fine, so long as you can appreciate the expediency of being able to ignore the velocity components entirely and dealing directly with the energy. It's certainly much speedier in an exam, for example.