The horizontal and vertical displacement $x$ and $y$ of a projectile at a given time $t$ are given by $x = 6t$ metre and $y = 8t - {5t}^{2}$ metre. The range of the projectile in metre is

Question

Dr. Abubacker · Accepted Answer

(a) $x = (u\cos{	heta)t = 6t}$
$y = (u\sin{	heta)\ t - \frac{1}{2}{gt}^{2} = 8t - {5t}^{2}}$
$Therefore,\ \ \ u\sin{	heta = 8}$
$u\cos{	heta = 6}$
$Range.\ \ R = \frac{u^{2}2\sin{2	heta}}{g}$
$= \frac{u^{2} 	imes 2\sin{	heta\cos	heta}}{g}$
$= \frac{2(u\sin{	heta)\ (\cos{	heta)}}}{g}$
$= \frac{2(8)(6)}{10} = 9.6\ m$

Q. The horizontal and vertical displacement $x$ and $y$ of a projectile at a given time $t$ are given by $x = 6t$ metre and $y = 8t - {5t}^{2}$ metre. The range of the projectile in metre is

Explanation

Reviewed by Dr. Abubacker

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