Q225 / 225 · JEE Main 2004 · Physics

If $t_1$ and $t_2$ are the times of flight of two particles having the same initial velocity $u$ and range $\mathrm{R}$ on the horizontal, then $\mathrm{t}_1^2+\mathrm{t}_2^2$ is equal to

  • A
    $\frac{u^2}{g}$
  • Correct answer
    B
    $\frac{4 u^2}{g^2}$
  • C
    $\frac{u^2}{2 g}$
  • D
    1

Solution

$\mathrm{t}_1=\frac{2 \mathrm{u} \sin \alpha}{\mathrm{g}}, \mathrm{t}_2=\frac{2 \mathrm{u} \sin \beta}{\mathrm{g}}$ where $\alpha+\beta=90^{\circ}$ $\therefore \mathrm{t}_1^2+\mathrm{t}_2^2=\frac{4 \mathrm{u}^2}{\mathrm{~g}^2}$

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

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