What is the Launch Angle of a Projectile at Half Its Maximum Height?

[hawkeye1029]

Homework Statement

At 0.5 of its maximum height, the speed of a projectile is 0.75 of its initial speed. What was its launch angle?

Homework Equations

Not sure here, but:
v^2 - vo^2 = 2as [maximum height]
Ay = Asin(theta) [y-component]

The Attempt at a Solution

I didn't get very far, but:
vo = vo
v = 0.75 vo

0.75vo^2 - vo^2 = (2)(-9.8)s
-0.25vo^2 = -19.6s
vo^2 = 784s

I'm stuck from here on, should I try to find the y-component for this?
[Also I'm an almost total beginner at Physics, excuse me if I'm totally wrong].

Any help would be greatly appreciated. Thanks everyone!

 

[hawkeye1029]

I thought about it a little more:

x-component : [v0^2/g]sin(theta)
y-component : vo sin(theta)

(theta) = tan^-1 [vo sin(theta)] / [vo^2/9.8 sin (theta)]
(theta) = tan^-1 (9.8/vo)

Is this headed in the right direction? If so, what would I substitute in for vo? If not, any tips?
Thanks!

 

[hawkeye1029]

Oh wait I think it would be best to use energy conservation.

0.5 mv^2 = 0.5m(0.75v)62 + 0.5mgh
and then
0.5 mv^2 = 0.5 m(v cos theta)^2 + 7/16(mv^2)
and simplify to get theta = 69 degrees

Is this correct?

 

[stockzahn]

Oh wait I think it would be best to use energy conservation.

0.5 mv^2 = 0.5m(0.75v)62 + 0.5mgh
and then
0.5 mv^2 = 0.5 m(v cos theta)^2 + 7/16(mv^2)
and simplify to get theta = 69 degrees

Is this correct?

I got the same value for θ₀ (but I don't really understand your way to find it). So I think the answer is correct, but the calculation is confusing.