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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.




$v_0 = g * t/2$

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

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

I used wolframalpha to solve for $\theta$.

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


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

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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. –  Will Cross Jan 18 '13 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. –  Duncan Murray Jan 18 '13 at 4:46
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1 Answer

up vote 2 down vote accepted

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:

$$t = 2 \frac{v_y}{g}$$ $$h = \frac{v_y^2}{2g}$$

Your horizontal range requires knowing the horizontal velocity, which you cannot figure out from the data you have.

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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 $v_y$. –  Duncan Murray Jan 19 '13 at 4:22
Nope, if the vertical velocity stays the same, horizontal velocity is irrelevant to time in the air and maximum height reached. If you keep the total velocity constant and change the angle from the vertical, you will increase range, at least until you reach 45°, but will reduce height and time in the air. –  Jaime Jan 19 '13 at 4:28
I think I understand now. Thanks. Now to figure out another way to estimate horizontal velocity. –  Duncan Murray Jan 19 '13 at 5:16
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