The horizontal and vertical displacements x a | vedclass

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(A) Given,$x = 6t$ and $y = 8t - 5t^2$.Comparing these with the standard equations of projectile motion:$x = (u \cos 	heta)t \implies u \cos 	heta =

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$9.6$

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(A) Given,$x = 6t$ and $y = 8t - 5t^2$.Comparing these with the standard equations of projectile motion:$x = (u \cos 	heta)t \implies u \cos 	heta = 6 	ext{ m/s}$.$y = (u \sin 	heta)t - \frac{1}{2}gt^2 = 8t - 5t^2$.Comparing the coefficients,we get $u \sin 	heta = 8 	ext{ m/s}$ and $\frac{1}{2}g = 5$,which implies $g = 10 	ext{ m/s}^2$.The range $R$ of a projectile is given by the formula:$R = \frac{u^2 \sin 2	heta}{g} = \frac{2(u \sin 	heta)(u \cos 	heta)}{g}$.Substituting the values:$R = \frac{2 	imes 8 	imes 6}{10} = \frac{96}{10} = 9.6 	ext{ m}$.

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

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