When a particle is projected at same angle to the horizontal, it has a range R and time of flight t1. If the same particle is projected with the same speed at some other angle to have the same range, its time of flight is t2, then

Question

When a particle is projected at same angle to the horizontal, it has a range R and time of flight t1. If the same particle is projected with the same speed at some other angle to have the same range, its time of flight is t2, then (A) t1+t2=(2 R/g) (B) t1-t2=(R/g) (C) t1 t2=(2 R/g) (D) t1 t2=(R/g)

Tardigrade Admin · Accepted Answer

The angles has to be complimentary i.e., if θ1 arrow θ, θ2 arrow(90-θ) t1=(2 u sin θ/g), t2=(2 u sin (90-θ)/g) t2=(2 u cos θ/g) t1 t2=(2 u sin θ/g) × (2 u cos θ/g) t1 t2-(2 R/g)

Q. When a particle is projected at same angle to the horizontal, it has a range $R$ and time of flight $t_{1}$ . If the same particle is projected with the same speed at some other angle to have the same range, its time of flight is $t_{2}$ , then

$t_{1} + t_{2} = \frac{2 R}{g}$

$t_{1} - t_{2} = \frac{R}{g}$

$t_{1} t_{2} = \frac{2 R}{g}$

$t_{1} t_{2} = \frac{R}{g}$

The angles has to be complimentary
i.e., if $θ_{1} \to θ, θ_{2} \to (90 - θ)$
$t_{1} = \frac{2 u s i n θ}{g},$
$t_{2} = \frac{2 u s i n ( 90 - θ )}{g}$
$t_{2} = \frac{2 u c o s θ}{g}$
$t_{1} t_{2} = \frac{2 u s i n θ}{g} \times \frac{2 u c o s θ}{g}$
$t_{1} t_{2} - \frac{2 R}{g}$

Q. When a particle is projected at same angle to the horizontal, it has a range R and time of flight t1​. If the same particle is projected with the same speed at some other angle to have the same range, its time of flight is t2​, then

t1​+t2​=g2R​

t1​−t2​=gR​

t1​t2​=g2R​

t1​t2​=gR​