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The horizontal and vertical displacements $x$ and $y$ of a projectile at a given time $t$ are given by $x=6 t$ metre and $y=8 t-5 t^2$ metre. The range of the projectile in metre is
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Verified Answer
The correct answer is:
$9.6$
$$
\begin{aligned}
& x=(u \cos \theta) t=6 t \\
& \quad y=(u \sin \theta) t-\frac{1}{2} g t^2=8 t-5 t^2
\end{aligned}
$$
Therefore, $u \sin \theta=8$
$u \cos \theta=6$
$\begin{aligned} Range\ R & =\frac{u^2 \sin 2 \theta}{g}=\frac{u^2 \times 2 \sin \theta \cos \theta}{g} \\ & =\frac{2(u \sin \theta)(u \cos \theta)}{g} \\ & =\frac{2(8)(6)}{10}=9.6 \mathrm{~m}\end{aligned}$
\begin{aligned}
& x=(u \cos \theta) t=6 t \\
& \quad y=(u \sin \theta) t-\frac{1}{2} g t^2=8 t-5 t^2
\end{aligned}
$$
Therefore, $u \sin \theta=8$
$u \cos \theta=6$
$\begin{aligned} Range\ R & =\frac{u^2 \sin 2 \theta}{g}=\frac{u^2 \times 2 \sin \theta \cos \theta}{g} \\ & =\frac{2(u \sin \theta)(u \cos \theta)}{g} \\ & =\frac{2(8)(6)}{10}=9.6 \mathrm{~m}\end{aligned}$
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