Question

A projectile can have the same range $R$ for two angles of projection. if $t_1$ & $t_2$ be the times of flight in the two cases then what is the product of the two times of flight?

Answer 1

Step 1: Formulas used

If $R$ is the range of a projectile thrown at an angle of $\theta$, then the range of a projectile thrown at an angle of $90°-\theta$ is also $r$.

We know that,

$t_1=\frac{2u\sin \theta }{g}$ and

$t_2=\frac{2u\sin \left(90°-\theta \right)}{g}=\frac{2u\cos \theta }{g}$

We also know that,

$R=\frac{u^2\sin \left(2\theta \right)}{g}$

Where, $u$ is the initial velocity of the projectile and $g$ is the acceleration due to gravity.

Step 2: Calculate the product of the two times of flight

Multiplying $t_1$ and $t_2$,

$\begin{array}{rcl}{t}_1·t_2& =& \frac{2u\sin \theta }{g}·\frac{2u\cos \theta }{g}\\ & =& \frac{4u^2\sin \theta \cos \theta }{g^2}\\ & =& \frac{2\left[{\displaystyle \frac{u^2\sin \left(2\theta \right)}{g}}\right]}{g}\\ \therefore t_1·t_2& =& \frac{2R}{g}\end{array}$

Therefore, the product of the two time of flights $t_1$ and $t_2$ is $\frac{2R}{g}$.