How to Solve for Initial Velocity and Angle?

Question

I hope you're having a pleasant day. Wondering if I could ask for your assistance on some projectile problem in 2d plane.

How do I get the initial velocity and angle if maximum height, flight time, initial height, final height, and range are already given?

Basically I just wanted to compute for the force needed and the angle with the given parameters. The initial height or the initial elevation of the projectile, and the final height or the target elevation can be exchanged -- what I mean by this is the instances where one of them is on the higher ground.

Please also note that I'm very new to interpreting math equations, a step-by-step breakdown of the solution would help me greatly. Very sorry for the trouble.

Please let me know if my question is unclear. Thank you in advance!

Answer 1

This is a typical orbit determination problem. The orbit kinematics for this problem gives

$$ \cases{ y = y_i + v_0 \sin\theta t-\frac 12 g t^2\\ x = x_i + v_0 \cos\theta t } $$

now the known numbers give us

$$ \cases{ y_f = y_i + v_0 \sin\theta t_f-\frac 12 g t_f^2\\ x_f = x_i + v_0 \cos\theta t_f } $$

also the maximum height condition $\dot y = v_0\sin\theta - g t^* = 0$ gives us

$$ t^* = \frac{v_0\sin\theta}{g} $$

and then

$$ y_{max} = y_i + \frac{v_0^2\sin^2\theta}{g}-\frac 12\frac{v_0^2\sin^2 \theta}{g} = y_i +\frac 12\frac{v_0^2\sin^2 \theta}{g} $$

also we have

$$ d = x_f - x_i = v_0 \cos\theta t_f $$

Now if you know $y_i, y_f, y_{max}, t_f, d$ you can easily obtain $v_0, \theta, x_f$ assuming $x_i = 0$

Answer 2

Let $v$ be the initial speed, $\theta$ the initial angle to the horizontal. Let $d$ be the range (ie total horizontal distance travelled. Let $H$ be the maximum height (above the initial height).

The horizontal velocity is $v\cos\theta$ throughout the flight, so $d=Tv\cos\theta$. Hence $v^2\cos^2\theta=d^2/T^2$.

We have $H=v^2\sin^2\theta/(2g)$, so $$v^2=2gH+d^2/T^2$$ Then we get $\theta$ from $d=Tv\cos\theta$.