Question

A particle is projected with speed $u$ to just pass over a tower of height $h$. The product of the two possible times taken to pass over the tower is

• A. $\frac{2u}{g}$
• B. $\frac{2h}{g}$
• C. $\frac{u}{g}$
• D. None

Answer 1

The correct option is B $\frac{2h}{g}$

Taking upward direction as positive, the equation of trajectory in vertical direction will be,
$y=u\sin \alpha -\frac{1}{2}g{t}^2$
where $\alpha$ is the angle of projection.
Or $\frac{1}{2}g{t}^2-u\sin \alpha +y=0$
This is a quadratic equation in time with two roots $t_1$ and $t_2$

$t_1\times t_2=\frac{y}{\frac{1}{2}g}=\frac{2y}{g}(\text{Since}y=h)\therefore t_1\times t_2=\frac{2h}{g}$