Finding Launch Angle Using Projectile Motion Equations

[Imuell1]

Homework Statement

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.)

Homework Equations

The Attempt at a Solution

I don't even know where to begin.

 

[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?

 

[Imuell1]

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

 

[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| = (V_x² + V_y²)^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.

 

[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.

 

[LowlyPion]

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

Isn't it

V_h/2² = V_o² -2*g*H/2 ?

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

V_h² = 0 = V_o² -2*g*H

So ... substituting in the first one you get

V_h/2² = 2*g*h -2*g*H/2 = 2*g*H/2 = (V_o²)/2

That is a very important result.

Now look at my initial hint ...

 

[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?

 

[LowlyPion]

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?

What I did was take the additional fact that |V_h/2| = 3/4 |V_o|

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

Since V_h/2² = V_x² + V_yh/2² = (3/4*V_o²)

Just expand using the knowledge that at H/2 V_y² = 2*V_yh/2²