Question

It is possible to project a particle with a given speed in two possible ways so that it has the same horizontal range $R$. The product of the times taken by it in the two possible ways is

• A. $\frac{R}{g}$
• B. $\frac{2R}{g}$
• C. $\frac{3R}{g}$
• D. $\frac{4R}{g}$

Answer 1

Answer: (B)

$\frac{2R}{g}$

Range, $R=\frac{v_o^2\sin 2\theta }{g}$
since range is equal, $R_1=R_2$
$\because$ velocity is equal, $v_1=v_2$

$\Rightarrow \frac{v_1^2\sin 2{\theta }_1}{g}=\frac{v_1^2\sin 2{\theta }_2}{g}\Rightarrow \sin 2{\theta }_1=\sin 2{\theta }_2$
$\Rightarrow 2{\theta }_1={180}^{\circ }-2{\theta }_2\Rightarrow {\theta }_1={90}^{\circ }-{\theta }_2$....(i)

Time of height, $T=\frac{2v_o\sin \theta }{g}$
product of two time of flight having same range,

$T_1\times T_2=\frac{2v_0\sin {\theta }_1}{g}\times \frac{2v_o\sin {\theta }_2}{g}$

$=\frac{4v_o^2\sin {\theta }_1\sin {\theta }_2}{g^2}$

from equation (i)
$=2\times \frac{v_o^{2}2\sin {\theta }_1\times \sin (90-{\theta }_1)}{g^2}$

$=\frac{2}{g}\times \frac{v_o^{2}2\sin {\theta }_1\cos {\theta }_1}{g}$

$\Rightarrow T_1\times T_2=\frac{2}{g}\frac{v_o^2\sin 2{\theta }_1}{g}=\frac{2R}{g}$
Hence, option B is correct