Question #a2924

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Question #a2924

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Answer 1 · 2014-12-11T13:17:57

The general equation of a parabola is as follows.

$y = \frac{- g x^{2}}{2 V^{2} {cos}^{2} A} + x tan A$ $y = (-gx^2)/(2V^2Cos^2 A) + xtan A$

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

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

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

$x = \frac{- g x^{2}}{2 V^{2} {cos}^{2} A} + x tan A$ $x = (-gx^2)/(2V^2cos^2 A) + xtan A$ ,

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

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

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

$x = \frac{(\frac{sin A}{cos A} - 1) (2 V^{2} {cos}^{2} A)}{g}$ $x = [((sin A)/(cos A) - 1)(2V^2 cos^2 A)]/g$

$x = \frac{(2 V^{2} sin A cos A - 2 V^{2} {cos}^{2} A)}{g}]$ $x = [(2V^2sin A cos A - 2V^2 cos^2 A)]/g]$

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

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

$x = \frac{V^{2} (sin 2 A - 2 {cos}^{2} A)}{g}$ $x =[ V^2( sin 2A - 2cos^2 A)]/g$

then d = rt

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

so

$t = \frac{V (sin 2 A - 2 {cos}^{2} A)}{g}$ $t = [V( sin 2A - 2 cos^2A)]/g$

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

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Question #a2924

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