For a given velocity, a projectile has the same range R for two angles of projection if t1 and t2 are the times of flight in the two cases then

The correct option is t1 t2 ∝ R.

Step-by-Step Explanation:

For a projectile projected with a given initial velocity
$u$
at an angle
$\theta$
with the horizontal, the horizontal range
$R$
is the same for two complementary projection angles:
$\theta$
and
$(90°-\theta )$ .

Let's write down the formulas for the time of flight for both projection angles.

For the first projection angle
$t_1=\frac{2usin\theta }{g}$
where
$g$
is the acceleration due to gravity.

For the second projection angle
$t_2=\frac{2usin(90°-\theta )}{g}=\frac{2ucos\theta }{g}$ .

Now, let's find the product of the two times of flight,
$t_1{t}_2$ :
$t_1{t}_2=\left(\frac{2usin\theta }{g}\right)·\left(\frac{2ucos\theta }{g}\right)$
$t_1{t}_2=\frac{2u^2(2sin\theta cos\theta )}{g^2}$
Using the trigonometric identity
$2sin\theta cos\theta =sin2\theta$ , we get:
$t_1{t}_2=\frac{2}{g}\left(\frac{u^{2}sin2\theta }{g}\right)$ .

Since the horizontal range of the projectile is given by
$R=\frac{u^{2}sin2\theta }{g}$ , we can substitute
$R$
into our equation:
$t_1{t}_2=\frac{2R}{g}$ .

Since
$g$
is a constant, the product of the times of flight is directly proportional to the horizontal range:
$t_1{t}_2\propto R$ .