Question

When a particle is projected at same angle to the horizontal, it has a range $R$ and time of flight $t_{1}$. If the same particle is projected with the same speed at some other angle to have the same range, its time of flight is $t_{2}$, then

• A. $t_{1}+t_{2}=\frac{2 R}{g}$
• B. $t_{1}-t_{2}=\frac{R}{g}$
• C. $t_{1} t_{2}=\frac{2 R}{g}$
• D. $t_{1} t_{2}=\frac{R}{g}$

Answer 1

Answer: (C)

The angles has to be complimentary
i.e., if $\theta_{1} \rightarrow \theta, \theta_{2} \rightarrow(90-\theta)$
$t_{1}=\frac{2 u \sin \theta}{g},$
$t_{2}=\frac{2 u \sin (90-\theta)}{g}$
$t_{2}=\frac{2 u \cos \theta}{g}$
$t_{1} t_{2}=\frac{2 u \sin \theta}{g} \times \frac{2 u \cos \theta}{g}$
$t_{1} t_{2}-\frac{2 R}{g}$