# Finding Angle of Trajectory

• Lamebert
In summary: I think I need to find an equation for θ that will hold for all h.In summary, the projectile is launched at angle θ0 with the horizontal and has half the speed it had when it was at half its maximum height h/2.

## Homework Statement

A projectile is launched from the earth’s surface at initial speed vi 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?

## The Attempt at a Solution

I know that the velocity vector at the highest point is vicosθ, and I know that the maximum height of the parabola is (visinθ)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?

Hi Lamebert! Welcome to PF!

You need to translate …
Lamebert said:
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

tiny-tim said:
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.

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

sooo the equation to use is … ?

tiny-tim said:
well, you know the velocity at two different points, and the distance between them (and the acceleration),

sooo the equation to use is … ?