# Find the launch angle

1. Feb 11, 2009

### Imuell1

1. The problem statement, all variables and given/known data

at 1/2 of its maximum height, the speed of a projectile is 3/4 of its initial speed. what was its launch angle?(ignore any effects due to air resistance.)

2. Relevant equations

3. The attempt at a solution

I don't even know where to begin.

2. Feb 11, 2009

### NBAJam100

did you try looking at the projectile motion equations and seeing how you can work with them knowing that you can put your velocity at half of its max height in terms of the initial velocity?

3. Feb 11, 2009

### Imuell1

no i didn't try that. i am still so confused. which equation would i start with?

4. Feb 12, 2009

### LowlyPion

This problem is a little tricky, but not too much so, if you focus on velocity and speed.

Keep in mind what speed is at all points. It is the |velocity|.

At any point |V| = (Vx2 + Vy2)1/2

Then all you have to do is figure the velocity in the vertical direction when it is gone half way and compare the |velocities| (speed) according to what the problem asks.

5. Feb 12, 2009

### Imuell1

Ugh, I am still so confused. I emailed my professor also and he said to use the equation Vy^2-Vyo^2=2ay and to put H in terms of Vy. For some reason I still have no idea where to begin still. I got y=(Vy^2-Vyo^2)/2a but even if that is correct, I don't know what to do next.

6. Feb 12, 2009

### LowlyPion

By all means that is the correct equation to use. At the H/2 point how fast is Vy?

Isn't it

Vh/22 = Vo2 -2*g*H/2 ?

And you know that at H (which is max height)

Vh2 = 0 = Vo2 -2*g*H

So ... substituting in the first one you get

Vh/22 = 2*g*h -2*g*H/2 = 2*g*H/2 = (Vo2)/2

That is a very important result.

Now look at my initial hint ...

7. Feb 13, 2009

### Imuell1

when y=h, H=-Vyo^2/2a

so then to find the velocity of h/2 do i make the equation
H/2=(Vy^2-Vyo^2)/2a ?

then i substitute for H and get
(-Vyo^2/2a)/2=(Vy^2-Vyo^2)/2a?

8. Feb 13, 2009

### LowlyPion

What I did was take the additional fact that |Vh/2| = 3/4 |Vo|

Take advantage of the fact that Vx is constant through out.

Since Vh/22 = Vx2 + Vyh/22 = (3/4*Vo2)

Just expand using the knowledge that at H/2 Vy2 = 2*Vyh/22