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 ie,

θ

and

(9 0^{\circ} - θ)

.

In both cases

t_{1} = \frac{2 u s i n θ}{g}

... (i)

t_{2} = \frac{2 u s i n ( 9 0 ^{\circ} - θ )}{g}

= \frac{2 u c o s θ}{g}

... (ii)
From Eqs. (i) and (ii)

t_{1} t_{2} = \frac{4 u ^{2} s i n θ c o s θ}{g ^{2}}

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

= \frac{2}{g} \frac{u ^{2} s i n 2 θ}{g}

∴ t_{1} t_{2} = \frac{2 R}{g}

(∴ R = \frac{u ^{2} s i n 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

$R^{2}$

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

$\frac{1}{R}$

$R$