Projectile Launch Angle Calculation with Given Launch Speed and Maximum Height

[shell4987]

Homework Statement

A projectile's launch speed is 6.4 times its speed at maximum height. Find the launch angle 0.

Homework Equations

I know that the x-axis has to be the horizontal and the y-axis, vertical. I know I also have to use the square root of V0x squared plus V0y squared = 6.4 (V0x) and that I need to solve for V0y and then the inverse tangent to find the launch angle.

The Attempt at a Solution

I tried letting V0x and V0y be the components of the initial velocity. Also, at max height, the speed of the projectile is V0x since it is traveling horizontally and that component of velocity is constant. I don't really know where to go from there because I don't know what to put for the V0x... I know how to figure out this problem, but I just don't know what V0x is or how to figure it out, does anyone have any clue on how to figure that out?

 

[learningphysics]

you're almost there...

square both sides... solve for Vx, in terms of Vy... then take Vy/Vx, then inverse tan.

 

[shell4987]

thank you SO much, i got the answer to be 81.1 degrees :)