Projectiles Launched at an Angle with unspecified Initial Velocity I'm attempting to do my Physics homework, and I did the first one right, but that problem gave me initial velocity. This problem gives me only the angle relative to horizontal and the distance it covers. Can anyone help me figure out where to start? I've tried but I can't find any formula that I can find initial velocity without having time, or vice versa. Any help would be much appreciated. Here's the problem in full: 

A golfer hits a golf ball at an angle of 25.0° to the ground. If the golf ball covers a horizontal distance of 301.5 meters, what is the ball's maximum height? (Hint: At the top of its flight, the ball's vertical velocity component will be zero.) 

I realize that the vertical velocity component has something to do with it, but I can't figure out where that would fit in. 
 A: In projectile motion the horizontal velocity is always same through the journey, only vertical component of velocity changes. 
After resolving the given velocity vector U (say) into X and Y components as Ux and Uy respectively. 
You can write 
R=Ux.T ( R is horizontal range, T is time of flight) 
Therefore, R=u^2 sin(2 theta)/g…(1)
Here theta is the angle between velocity vector with the horizontal. 
You can derive the expression for maximum height using 3rd equation of motion -
H=u^2sin^2(theta)/2g…(2)
Here H is maximum height. 
Now combining equations (1) & (2)
We get 
H=R.tan(theta)/4
You can use this relation for deriving the maximum height of the projectile if Range and angle is given.
A: First, you have to find the maximum height of the ball. In order to do so, we use the condition that $v_{y}=0$. 
$$0=v_{0}\sin(\alpha)-gt_{H}$$
From this, the time to reach the maximum height $t_{H}$ is 
$$t_{H}=\frac{v_{0}\sin(\alpha)}{g}$$
From the vertical displacement, we obtain the maximum height using $t_{H}$
$$H=\frac{v_{0}^{2}\sin^{2}(\alpha)}{2g}$$
But you dont know the initial velocity. Only the angle and the range. You find the range by setting $y=0$
$$0=v_{0}\sin(\alpha)t_{D}-\frac{gt_{D}^{2}}{2}$$
Following the same steps as above, you get the time of travel $t_{D}$ and from this you find the range to be 
$$D=\frac{v_{0}^{2}\sin(2\alpha)}{g}$$
From all this you get
$$\frac{H}{D}=\frac{\sin^{2}(\alpha)}{2\sin(2\alpha)}\Longrightarrow H=\frac{D\tan(\alpha)}{4}$$
