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Physics Question on Motion in a plane

For a given velocity, a projectile has same range RR for two angles of projections. If t1t_1 and t2t_2 are the time of flight in the 2 cases then

A

t1t2∝1R2t_1t_2\propto\frac{1}{R^2}

B

t1t2∝1Rt_1t_2\propto\frac{1}{R}

C

t1t2∝Rt_1t_2\propto R

D

t1t2∝R2t_1t_2\propto R^2

Answer

t1t2∝Rt_1t_2\propto R

Explanation

Solution

R = u2 sin 2θ1g\frac{u^2 \, sin \, 2\theta_1}{g}
=u2 sin 2θ2g= \frac{u^2 \, sin \, 2\theta_2}{g}
Also, θ1+θ2=90∘\theta_1 + \theta_2 = 90^\circ
i.e. θ2=90∘−θ1\theta_2 = 90^\circ - \theta_1
Also, t1=2u sin θ1gt_1 = \frac{2u\, sin \, \theta_1}{g} and
t2=2u sin θ2gt_2 = \frac{2u\, sin \, \theta_2}{g}
then
t1 t2=4u2g2 sin θ1 sin θ2t_1 \,t_2 = \frac{4u^2}{g^2} \, sin \, \theta_1 \, sin \, \theta_2
i.e.
t1 t2=2gu2g2 sin θ1 sin(90−θ1)t_1 \, t_2 = \frac{2}{g} \frac{u^2}{g} 2 \, sin \, \theta_1 \, sin (90 - \theta_1)
=2gu2g2 sin θ1 cos θ1= \frac{2}{g} \frac{u^2}{g} 2 \, sin \, \theta_1 \, cos \, \theta_1
=2gu2 sin 2θ1g= \frac{2}{g} \frac{u^2 \, sin \, 2\theta_1}{g}
=2gR= \frac{2}{g}R
i.e. t1 t2∝Rt_1 \, t_2 \propto R