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In my studies regarding projectile motion, I recently completed a simple level question surrounding a multileveled golf range. I had to determine the trajectories of these balls after they were shot into the air.

I was wondering if some sort of mathematical model could be developed for finding the horizontal distance, d in terms of v (initial velocity), h (initial height) and θ (angle to the horizontal) from each of the three levels, if the bottom level was treated as 0, the middle level, h, and the top level 2h.

Thankyou

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    $\begingroup$ Welcome to MSE. What have you tried? Can you at least update your question with the equations of the projectile? $\endgroup$ – mathcounterexamples.net May 2 at 8:55
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If a projectile starts out at position $(0,h)$, initial velocity $v$ and angle to the horizontal $\theta$ then at time $t$ it has co-ordinates $(x(t),y(t))$ where

$x(t) = vt \cos \theta \\ y(t) = h + vt \sin \theta - \frac 1 2 gt^2$

We can eliminate $t$ and write the equation of the projectile's path as

$y = h + x \tan \theta - \frac{g}{2v^2 \cos^2 \theta}x^2$

Set $y=0$ and you are left with a quadratic equation in $x$. Solve for $x$ and you have the two values of $x$ when the projectile is at ground level. Only one of these values is positive, so that is the one you want.

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A good way of thinking about this is thinking about the time the ball is in the air.

We know that $v_x = v\cos\theta$, so multiplying this by the total 'air time' will give the range of the projectile.

For our hieght, we have $y=-\frac{g}{2}t^2 + vt\sin\theta+y_0$, where $y_0$ is the initial height.

To get the total flight time, we simply substitute $y=0$ (where the projectile will hit the ground), and solve for $t$ using quadratic formula. Then multiplying by $v_x$ as mentioned before will give the range.

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