For a given velocity, a projectile has same range R for two angles of projections. If t1 and t2 are the time of flight in the 2 cases then

Question

For a given velocity, a projectile has same range R for two angles of projections. If t1 and t2 are the time of flight in the 2 cases then (A) t1t2 propto(1/R2) (B) t1t2 propto(1/R) (C) t1t2 propto R (D) t1t2 propto R2

Tardigrade Admin · Accepted Answer

R = (u2 sin 2θ1/g) = (u2 sin 2θ2/g) Also, θ1 + θ2 = 90° i.e. θ2 = 90° - θ1 Also, t1 = (2u sin θ1/g) and t2 = (2u sin θ2/g) then t1 t2 = (4u2/g2) sin θ1 sin θ2 i.e. t1 t2 = (2/g) (u2/g) 2 sin θ1 sin (90 - θ1) = (2/g) (u2/g) 2 sin θ1 cos θ1 = (2/g) (u2 sin 2θ1/g) = (2/g)R i.e. t1 t2 propto R

Q. For a given velocity, a projectile has same range $R$ for two angles of projections. If $t_{1}$ and $t_{2}$ are the time of flight in the 2 cases then

$t_{1} t_{2} \propto \frac{1}{R ^{2}}$

$t_{1} t_{2} \propto \frac{1}{R}$

$t_{1} t_{2} \propto R$

$t_{1} t_{2} \propto R^{2}$

Q. For a given velocity, a projectile has same range R for two angles of projections. If t1​ and t2​ are the time of flight in the 2 cases then

t1​t2​∝R21​

t1​t2​∝R1​

t1​t2​∝R

t1​t2​∝R2

Q. For a given velocity, a projectile has same range $R$ for two angles of projections. If $t_{1}$ and $t_{2}$ are the time of flight in the 2 cases then

$t_{1} t_{2} \propto \frac{1}{R ^{2}}$

$t_{1} t_{2} \propto \frac{1}{R}$

$t_{1} t_{2} \propto R$

$t_{1} t_{2} \propto R^{2}$