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.
Best Answer
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.