A projectile can have the same range R for two angles of projection. If t1 and t2 be the times of flights in the two cases, then the product of the two times of flights is proportional to :

Question

A projectile can have the same range R for two angles of projection. If t1 and t2 be the times of flights in the two cases, then the product of the two times of flights is proportional to : (A) R2 (B) (1/R2) (C) (1/R) (D) R

Tardigrade Admin · Accepted Answer

A .projectile can have same range if angles of projection are complementary i.e., 0 and

(9 0^{°} - θ)

. Thus, in both cases :

t_{1} = \frac{2 u s in θ}{g} ... (i)

t_{2} = \frac{2 u s in ( 9 0 ^{°} - θ )}{g}

= \frac{2 u cos θ}{g} ... (ii)

From Eqs.

(i)

and

(ii)

t_{1} t_{2} = \frac{4 u ^{2} s in θ cos θ}{g ^{2}}

t_{1} t_{2} = \frac{2 u ^{2} s in 2 θ}{g ^{2}}

= \frac{2}{g} \frac{u ^{2} s in 2 θ}{g}

∴ t_{1} t_{2} = \frac{2 R}{g} (∴ R = \frac{u ^{2} s in 2 θ}{g})

Hence,

t_{1} t_{2} \propto R

Q. A projectile can have the same range $R$ for two angles of projection. If $t_{1}$ and $t_{2}$ be the times of flights in the two cases, then the product of the two times of flights is proportional to :

$R^{2}$

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

$\frac{1}{R}$

$R$

Q. A projectile can have the same range R for two angles of projection. If t1​ and t2​ be the times of flights in the two cases, then the product of the two times of flights is proportional to :

R2

R21​

R1​

R

Q. A projectile can have the same range $R$ for two angles of projection. If $t_{1}$ and $t_{2}$ be the times of flights in the two cases, then the product of the two times of flights is proportional to :

$R^{2}$

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

$\frac{1}{R}$

$R$