Q155 / 160 · TS EAMCET 2014 · Physics

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 $=g$ )

  • A
    $\frac{g^2 T}{2}$
  • B
    $\frac{g T}{2}$
  • Correct answer
    C
    $\frac{g T^2}{2}$
  • D
    $\frac{g^2 T^2}{2}$

Solution

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 $$ \begin{aligned} & R_{\max }=\frac{1}{g}\left(\frac{T g}{\sqrt{2}}\right)^2 \\ & R_{\max }=\frac{1}{2} g T^2 \end{aligned} $$

Hence, the correct answer is option: 'C'.

Relevant Questions