So after decomposing the velocity vector I got:
Vi = -yf-at2 /t
Plugging that into the equation for h...
h= ((-yf-at2)2*(sin2θ)) / 2g
and then
0.5h = 0.31((-yf-at2)2*(sin2θ)) / 2g
Am I on the right track?
Homework Statement
The speed of a projectile when it reaches its maximum height is 0.31 times its speed when it is at half its maximum height. What is the initial projection angle of the projectile?
Homework Equations
h=(Vi^2 * sin^2 θ) / 2g
R=(Vi^2 * sin2θ) / g
The Attempt at a Solution...