# Question a2924

Dec 11, 2014

The general equation of a parabola is as follows.

$y = \frac{- g {x}^{2}}{2 {V}^{2} C o {s}^{2} A} + x \tan A$

$A$ is the angle at which the projectile is fired.
$g$ is $9.81 \frac{m}{s} ^ 2$, acceleration due to gravity.
$V$ the initial velocity

when $y = 0$ we have $x$ as the maximum range

To get the answer for this question, we set $x = y$ Therefore,

$x = \frac{- g {x}^{2}}{2 {V}^{2} {\cos}^{2} A} + x \tan A$,

then we divide both sides by $x$ then we have,

$1 = \frac{- g x}{2 {V}^{2} {\cos}^{2} A} + \tan A$

$g \frac{x}{2 {V}^{2} {\cos}^{2} A} = \tan A - 1$

$x = \frac{\left(\frac{\sin A}{\cos A} - 1\right) \left(2 {V}^{2} {\cos}^{2} A\right)}{g}$

 x = [(2V^2sin A cos A - 2V^2 cos^2 A)]/g]#

$2 \sin A \cos A = \sin 2 A$

$x = \frac{{V}^{2} \sin 2 A - 2 {V}^{2} {\cos}^{2} A}{g}$

$x = \frac{{V}^{2} \left(\sin 2 A - 2 {\cos}^{2} A\right)}{g}$

then d = rt

to get the time, t, we divide x with V

so

$t = \frac{V \left(\sin 2 A - 2 {\cos}^{2} A\right)}{g}$

I will still update this answer. If you have something to add pls do so. Thanks