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A body is projected at an angle $\theta$ so that its range is maximum. If $T$ is the time of flight, then the value of maximum range is (acceleration due to gravity $=g$ )
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Verified Answer
The correct answer is:
$\frac{g T^2}{2}$
We know that,
Range of projectile
$$
R=\frac{u^2 \sin 2 \theta}{g}
$$
As range is maximum, $\theta=45^{\circ}$
$$
R_{\max }=\frac{u^2 \sin 2 \times 45}{g}=\frac{u^2}{g}
$$
Flight time of projectile,
$$
T=\frac{2 u \sin 45^{\circ}}{g}
$$
$$
\begin{aligned}
& =\frac{2 u}{\sqrt{2} \cdot g}=\frac{\sqrt{2} \cdot u}{g} \\
\text { or } \quad u & =\frac{T g}{\sqrt{2}}
\end{aligned}
$$
or
Putting these value of $u$ in Eq. (i), we get
$$
\begin{aligned}
& R_{\max }=\frac{1}{g}\left(\frac{T g}{\sqrt{2}}\right)^2 \\
& R_{\max }=\frac{1}{2} g T^2
\end{aligned}
$$
Range of projectile
$$
R=\frac{u^2 \sin 2 \theta}{g}
$$
As range is maximum, $\theta=45^{\circ}$
$$
R_{\max }=\frac{u^2 \sin 2 \times 45}{g}=\frac{u^2}{g}
$$
Flight time of projectile,
$$
T=\frac{2 u \sin 45^{\circ}}{g}
$$
$$
\begin{aligned}
& =\frac{2 u}{\sqrt{2} \cdot g}=\frac{\sqrt{2} \cdot u}{g} \\
\text { or } \quad u & =\frac{T g}{\sqrt{2}}
\end{aligned}
$$
or
Putting these value of $u$ in Eq. (i), we get
$$
\begin{aligned}
& R_{\max }=\frac{1}{g}\left(\frac{T g}{\sqrt{2}}\right)^2 \\
& R_{\max }=\frac{1}{2} g T^2
\end{aligned}
$$
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