Mechanics Help (1 Viewer)

Andy005 · Jul 28, 2018

A particle is projected with speed V and angle of elevation alpha from point O on the edge of a cliff of height h. When the particle hits the ground its path makes an angle arctan(2tan(alpha)) with the horizontal.

Screen Shot 2018-07-28 at 11.34.49 am.png

Why is tan(beta)=vertical velocity/horizontal velocity?

Thanks.

fan96 · Jul 28, 2018

If the direction of the projectile's velocity makes an angle $\theta$ with the horizontal, then it is true that:

$\begin{aligned} \tan \theta = \frac{\dot y}{\dot x} &= \frac{-gt + V \sin \alpha}{V \cos \alpha} \\ &= -\frac{gt}{V}\sec \alpha + \tan \alpha\end{aligned}$

At the point where the projectile hits the ground, $\theta = - \arctan 2 \tan \alpha$

(the velocity vector of the projectile is pointing downwards, so $\theta$ is in the fourth quadrant).

$\begin{aligned} \tan (- \arctan 2 \tan\alpha) &= -\frac{gt}{V}\sec \alpha + \tan \alpha \\ -2 \tan \alpha &= -\frac{gt}{V}\sec \alpha + \tan \alpha \end{aligned}$

Now:

$\begin{aligned} \pi - \beta &= \arctan 2 \tan \alpha \\ \tan(\pi-\beta) &= 2 \tan \alpha \\ -\tan \beta &= 2 \tan \alpha \\ \tan \beta &= -2 \tan \alpha \\ &= -\frac{gt}{V}\sec \alpha + \tan \alpha\\ &= \frac{\dot y}{\dot x}\end{aligned}$

InteGrand · Jul 28, 2018

Andy005 said:
Why is tan(beta)=vertical velocity/horizontal velocity?

Thanks.

$\noindent This is true in general (for any (smooth) flight path, at any point on the curve). If $\beta$ is the angle made by the curve to the positive $x$-axis at some point in time, then $\tan \beta = \frac{v_{y}}{v_{x}}$ at that time (where $v_{x} \equiv \dot{x}$ and $v_{y}\equiv \dot{y}$). This is because \color{red}{the slope of the tangent to the curve at that point is} $\color{red}{\tan \beta}$ \color{black} and also \color{blue}{the velocity vector at that point is tangent to the curve} \color{black} at that point. By decomposing the velocity vector into its horizontal and vertical components and using ``$\text{slope} = \frac{\text{rise}}{\text{run}}$'', \color{red}{the slope of the tangent is} $\color{red}{\frac{v_{y}}{v_{x}}}$\color{black}. So $\tan \beta = \frac{v_{y}}{v_{x}}$.$

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Mechanics Help (1 Viewer)

Andy005

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fan96

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InteGrand

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