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So given basic knowledge about projectiles, the equation for a projectile's initial horizontal speed $V_x$ can be expressed as $V\cos\theta$, $V$ being the initial velocity. Then the horizontal distance can be written as $\delta x=V\cos\theta t$, then rearranged to be $t=\delta x/V\cos\theta$.

then subbing the equation for $t$ into the formula for vertical distance gives $y=V\sin\theta t-1/2gt^2$ which is an expression for vertical distance in terms of the horizontal distance, initial velocity and theta.

How can I rearrange this so that $V$ becomes the subject in terms of $\theta$, horizontal distance and vertical distance?

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    $\begingroup$ For guidance in typesetting mathematics, see this page in the help center. Help center>Our Model (physics.stackexchange.com/help/notation) $\endgroup$
    – TheFool
    Feb 7, 2017 at 9:00
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    $\begingroup$ This one is also helpful on typesetting. $\endgroup$ Feb 7, 2017 at 9:28

3 Answers 3

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To relate velocity and $\theta $ to distance, there is a slightly better way to approach it.

First start with your y equation $$y = Vt \sin {\theta} - \frac {1}{2} g t^2$$ and realize that you can solve for total time if you set $y=0$ using the quadratic equation. You will get 2 values. One of them will be 0 or a negative value (if you started at $y >0$). The 0 or negative value represents when you start on the ground.

You can then use that $t$ to solve for total horizontal distance. You also know that the highest point will be right in between the two times when $y=0$ (for example if your two t values are $t=0$ and $t=6$ then $t_{ymax} = 3$).

You can substitute $t_{ymax} $ into the y equation to find the max height.

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  • $\begingroup$ Sorry for our question we have to eliminate time from the equation and just have it in terms of the other variables. Thank you so much though! $\endgroup$
    – Lisa Ban
    Feb 7, 2017 at 12:21
  • $\begingroup$ @LisaBan You can do all of this in one equation. It will just look messier (see the other answers). Depending on how you approach it you could use my answer as a template and then just substitute all the messy variables at the end. $\endgroup$
    – JMac
    Feb 7, 2017 at 12:30
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$$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} ut\cos \theta \\ ut\sin \theta-\dfrac{gt^2}{2}\ \end{pmatrix}$$

Substitute $t=\dfrac{x}{u\cos \theta}$ into $y$,

\begin{align*} y &= x\tan \theta-\frac{gx^2}{2u^2\cos^2 \theta} \\ u^2 &= \frac{gx^2}{2\cos^2 \theta(x\tan \theta-y)} \\ u &= x\sqrt{\frac{g}{x\sin 2\theta-2y\cos^2 \theta}} \end{align*}

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  • $\begingroup$ I don't think that y should be squared in the final answer. $\endgroup$
    – D. Ennis
    Feb 7, 2017 at 13:12
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Substituting $t=\frac{\delta{x}}{V\cos\theta}$ into $y=V\sin\theta t-\frac12gt^2$, we get $$y=V\sin\theta\frac{\delta{x}}{V\cos\theta}-\frac12g\frac{\delta{x}^2}{V^2\cos^2\theta}$$ $$y=\delta{x}\;\tan\theta-\frac12g\frac{\delta{x}^2}{V^2\cos^2\theta}$$ $$\frac12g\frac{\delta{x}^2}{V^2\cos^2\theta}=\delta{x}\;\tan\theta-y$$ $$\frac12g\frac{\delta{x}^2}{(\delta{x}\;\tan\theta-y)\cos^2\theta}=V^2$$ $$V=\sqrt{\frac12g\frac{\delta{x}^2}{(\delta{x}\;\tan\theta-y)\cos^2\theta}}$$

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  • $\begingroup$ Note that this answer and Ng Chung Tak's are the same except for simplification and the use of a trig identity. $\endgroup$
    – D. Ennis
    Feb 7, 2017 at 13:20

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