# Calculating projectile range from known maximum height and time traveled

I've been stuck on this problem for many hours and I think I'm onto the right solution but I'm uncertain about my math.

I've got a projectile that I know its maximum height and it's hang time and I need to figure out it's range. Is it possible to calculate this with so few variables?

At the moment I'm ignoring air resistance and trying to work with a perfect parabola. I can approximate the launch velocity from the time the ball spends in the air, with $9.8m/s^2 * dt/2$ calculating gravity's effect to reach the time the ball starts falling.

I'm still pretty confused with calculating the angle only knowing these variables. $v_0$ is what I calculated as the launch velocity from the known time to peak height. $g = 9.8 m/s^2$, $h$ is the known height

Using the maximum height formula, I've worked out the following.

$h=(v\sin\theta)^2/2g$

$2gh=(v_0\sin\theta)^2$

$\sqrt{gh}=v_0\sin\theta$

$v_0 = g * t/2$

$\sqrt{2gh}=(4.9 m/s^2 * t)\sin(\theta)$

$\sin\theta = \sqrt{(2gh}/(g*t/2)$

I used wolframalpha to solve for $\theta$.

$\sin(\theta) = 2\sqrt{2}*\sqrt{gh}/gt$

$\theta=\sin^{-1}(2\sqrt{2}*\sqrt{gh}/gt)$

Am I on the right path with this? Any ways I can factor in wind resistance to this? Any help would be greatly appreciated.

• Don't factor in wind resistance; it gets annoying. As for hang time, you should already know how to calculate hang time for a drop from a certain height. The real question is, did you calculate $v_{0x}$? With that & the time, you can see how far it goes. Jan 18, 2013 at 4:18
• Thanks, I'm planing on using real world sensors for this, so I'll eventually have to factor in wind resistance, but for now perfect conditions will do. I'm using $9.8m/s^2 * dt/2$ to calculate $v_0x$ which calculates the time between the launch and the time that the vertical velocity = 0 (half the flight time for now). I'm still assuming that I'll need the angle of launch to appropriate the range however. Jan 18, 2013 at 4:46

You do not have enough information. Time in the air, $t$, and maximum height, $h$, are both a function of the vertical launch velocity, $v_y$ only:
$$t = 2 \frac{v_y}{g}$$ $$h = \frac{v_y^2}{2g}$$
• Wouldn't the time in the air also be a function of horizontal velocity? As the angle changes, you can reach the same height but have wildly varying hang times and ranges as a result. As the projectile is stationary at launch and would have a predictable trajectory, I'm struggling to understand how they are only functions of $v_y$. Jan 19, 2013 at 4:22