Projectile motion: solving for variable angle

[foo_daemon]

Hi,

I've been muddling over this problem for a few days. I thought there would be a simple approach, but I'm having trouble reaching a solution. Here is the rundown:

We have a 2D projectile launcher. It hurls an object from the origin (0,0) at some initial velocity with magnitude $V_i$ at angle θ with respect to the ground. You will be given its initial velocity, and must choose the launch angle θ such that the projectile will hit a target at point x,H (that is, a point x distance away with height H) during its descent (i.e. anytime after the peak where $V_y = 0$ ). It is a reasonable assumption that θ will be in the range $0 \leq θ \leq \frac{\pi}{2}$.

I am trying to find an explicit formula for θ in terms of $V_i$ , $x$ , $H$, and the usual $g = -9.8 \frac{m}{s^2}$
$V_{yi} = sin \theta V_i$
$V_{xi} = cos \theta V_i$

We have the following solutions for t:
$V_{yi} t + \frac{1}{2} g t^2 -H = 0$
$V_{xi} t - x = 0$

Since we want the projectile to hit the point (x,H) at the same time, we set the equations equal to each other:

$\frac {1}{2}g t^2 + ( V_{yi} - V_{xi} ) t + (x - H) = 0$

Use the quadratic formula to find t. (I expanded $V_{xi}$ and $V_{yi}$ in the resulting formula so we have θ back in the equation ).

$t = \frac{-V_i ( sin \theta - cos \theta ) \pm \sqrt{{V_i}^2 ( 1 - sin{2 \theta} ) - 2 g (x - H)}}{g}$

I used trig substitution in the square-root to make a $sin^2 + cos^2 = 1$ and double angle formula for $2 sin( \theta ) cos ( \theta ) = sin( 2 \theta )$ .

I want to factor that nasty expression under the radical by 'completing the square', but I don't see how that is possible.

Also, I don't really want to solve for t ... that is, I don't care how long the projectile takes. All I care about is what angle it needs to be launched at so that it hits point (x,H) .

 

[nasu]

If you don't need t, then why solve for it?
You can eliminate t between the two equations and find an equation in the variable theta.
For example, find t from the second one and plug it into the first one.
Then you can write the cos in terms of tan and find a quadratic equation in tan(theta).

 

[foo_daemon]

Aha... yes, I didn't really want to solve for t, I just didn't see a way around it. Your approach is simple, and I'm dumbfounded I didn't realize it first ;)

However, it still seems to be an involved problem. Now we have:

$t = \frac{x}{V_0}$
Plugging this t into the formula for $V_y$, I get:
$tan ( \theta ) x + \frac{ g x^2 }{2 cos^2 ( \theta ) V_0^2 } - H = 0$

This is very close to an explicit formula for θ, but I can't find any trig identities that will help me simplify. I can convert the $\frac{1}{cos^2}$ into $sec^2$ , but... that doesn't really seem to help.

 

[nasu]

Aha... yes, I didn't really want to solve for t, I just didn't see a way around it. Your approach is simple, and I'm dumbfounded I didn't realize it first ;)

However, it still seems to be an involved problem. Now we have:

$t = \frac{x}{V_0}$
Plugging this t into the formula for $V_y$, I get:
$tan ( \theta ) x + \frac{ g x^2 }{2 cos^2 ( \theta ) V_0^2 } - H = 0$

This is very close to an explicit formula for θ, but I can't find any trig identities that will help me simplify. I can convert the $\frac{1}{cos^2}$ into $sec^2$ , but... that doesn't really seem to help.

1/cos^2=1+tan^2

 

[foo_daemon]

Hm, I still had problems finding the solution. I 'cheated' and used WolframAlpha to solve that last formula for theta.

Wolfram uses substitution and gives:

$\theta = tan^{-1} \left ( \frac{ -V_i^2 \pm \sqrt{-g^2 x^2 + 2 g H V_i^2 + V_i^4}}{gx} \right )$

WolframAlpha then states that after plugging the substitution back into the formula and running a check, it determines this solution is 'incorrect'. I plugged in some test numbers a few times, and the value for theta appears correct (consistent).