# 2009 F=MA Exam #6 (Conceptual Velocity Question)

1. Jan 29, 2013

### SignaturePF

1. The problem statement, all variables and given/known data
An object is thrown with a ﬁxed initial speed v0 at various angles α relative to the horizon. At some constant
height h above the launch point the speed v of the object is measured as a function of the initial angle α.
Which of the following best describes the dependence of v on α? (Assume that the height h is achieved, and
assume that there is no air resistance.)

The answer is that v is independent of alpha.

2. Relevant equations
v2 = (v_0cosθ)2 - 2gd
Δx = v_ocosθt

3. The attempt at a solution
Let the displacement be d, as mentioned in the problem.
Solving for v gives:
v = √(v_02cos2θ - 2gd)
It looks like the velocity depends on the angle θ, particularly as θ increases, the velocity at that point seems to decrease.

How/why is this wrong?

2. Jan 29, 2013

### cepheid

Staff Emeritus
Actually the equation is:

v2 = v02 - 2gh

where v is the speed (magnitude of the velocity) at height h, and v0 is the initial speed (at height 0). As you can see, the speed at height h only depends on the initial speed and on the height reached. The individual components of the velocity don't enter into it.

This equation can be derived from the conservation of energy.

(KE + PE )before = (KE + PE)after

(1/2)mv02 + mg(0) = (1/2)mv2 + mgh

mv2 = mv02 + 2mgh

v2 = v02 - 2gh

3. Jan 29, 2013

### SignaturePF

Oh ok, thanks, so I messed up the equation.

4. Jan 29, 2013

### cepheid

Staff Emeritus
If you want to show it using the kinematics equations and the individual components, pretending that you don't know about conservation of energy, it's only a couple of extra steps.

We always have:

$v^2 = v_x^2 + v_y^2$, where these are the speeds at height h. Agreed?

In the x direction, there is no acceleration, so we have:

$v_x^2 = v_{x0}^2 = v_0^2 \cos^2 \alpha =~\textrm{const.}$

In the y-direction, we have a constant acceleration of -g, so we can use the kinematic equation

$v_y^2 = v_{y0}^2 -2gh = v_0^2\sin^2\alpha - 2gh$

Again, you can derive this equation purely from kinematics without appealing to conservation of energy if you don't want to. Just start with the other two kinematics equations $v_y = v_{y0} - gt$ and $h = v_{y0}t - (1/2)gt^2$, and eliminate t.

So now, plugging in the expressions for $v_x^2$ and $v_y^2$ to the first equation, we get:

$v^2 = v_0^2 \cos^2 \alpha + v_0^2\sin^2\alpha - 2gh$
$= v_0^2(\cos^2\alpha + \sin^2\alpha) - 2gh$
$= v_0^2 - 2gh$

Et voila!