Projectile Motion Question (1 Viewer)

abonu16 · Jul 9, 2016

In a general projectile motion question, how do u get to that quadratic equation in tan(theta)?

DatAtarLyfe · Jul 9, 2016

Derive the equation for x and y displacement (varies depending on the question). Then simultaneous the two to eliminate time (t).
It should come out after that, just change the sec^2(theta) to 1 + tan^2(theta)
Edit: oh yeh, and to remove the x and y, just sub in the values given from the question

InteGrand · Jul 9, 2016

abonu16 said:
In a general projectile motion question, how do u get to that quadratic equation in tan(theta)?

$\noindent Note $x = Vt\cos \theta \Rightarrow t = \frac{x}{V \cos \theta}$. Since $y = Vt\sin \theta -\frac{1}{2}gt^2$, we have$

$\begin{align*}y &= V\times \frac{x}{V \cos \theta} \times \sin \theta -\frac{1}{2}g\left( \frac{x}{V \cos \theta}\right)^2 \\ &= x\tan \theta -\frac{gx^2}{2 V^2} \sec^2 \theta \\ &= x\tan \theta -\frac{gx^2}{2V^2}\left(1+\tan^2 \theta\right) \\ &= - \frac{gx^2}{2V^2} \tan^2 \theta + x\tan \theta - \frac{gx^2}{2V^2}. \end{align*}$

$\noindent Here we assumed the projectile started at the origin. For a general starting point $\left(x_0,y_0\right)$, just replace every $x$ and $y$ you see above with $x-x_0$ and $y-y_0$, respectively.$

Make a Difference – Donate to Bored of Studies!

Projectile Motion Question (1 Viewer)

abonu16

New Member

DatAtarLyfe

Booty Connoisseur

InteGrand

Well-Known Member

Users Who Are Viewing This Thread (Users: 0, Guests: 1)