Finding the Angle of Trajectory

[Lamebert]

Homework Statement

A projectile is launched from the earth’s surface at initial speed v_i at angle θ₀ with the
horizontal. When the projectile is at its maximum height h, it has half the speed it had
when it was at half its maximum height h/2

At what angle was the projectile launched?

Homework Equations

The Attempt at a Solution

I know that the velocity vector at the highest point is v_icosθ, and I know that the maximum height of the parabola is (v_isinθ)² / 2g. At this point, I don't really know where to go. I'm not sure how to relate height and velocity together with what I know. I was doing this earlier and I tried to go further, but I just ended up with a lot of variable/trig function couples that don't make sense. I could take a picture of all the work I've done, but you wouldn't be able to follow it. I barely know what I tried.

You don't really have to answer it for me, just what should the next step be?

 

[tiny-tim]

Hi Lamebert! Welcome to PF!

You need to translate …

When the projectile is at its maximum height h, it has half the speed it had
when it was at half its maximum height h/2

… into one of the standard constant acceleration equations

 

[Lamebert]

Hi Lamebert! Welcome to PF!

You need to translate …


… into one of the standard constant acceleration equations

Thanks for the welcome, and can you go a little further? I already knew I was going to need those equations, I just don't know how they fit together, how velocity and height at each point should relate inside of the equations.

 

[tiny-tim]

well, you know the velocity at two different points, and the distance between them (and the acceleration),

sooo the equation to use is … ? ​

 

[Lamebert]

well, you know the velocity at two different points, and the distance between them (and the acceleration),

sooo the equation to use is … ? ​

v_f² = v_i² + 2ad

The problem is finding the angle theta, though. I know that v_icosθ = v, right? But if I set that equal to what I solved for from the above equation, that is, v = √(gh /3), i end up with two variables again, which doesn't really help me.