A projectile can have the same range 'R' for two angles of projection. If T1 and T2 be the time of flights in the two cases, then the product of the two time of flights is directly proportional to

Question

A projectile can have the same range 'R' for two angles of projection. If T1 and T2 be the time of flights in the two cases, then the product of the two time of flights is directly proportional to (A) 1/R2 (B) 1/R (C) R (D) R2

Tardigrade Admin · Accepted Answer

Range is same for complimentary angles. T1 = (2u sin θ/g) and T2 = (2u sin(90-θ)/g) and R = (u2 sin 2θ/g) ∴ T1T2 = (2u sin θ/g) × (2u cos θ/g) = (2R/g)

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

$1/ R^{2}$

$1/ R$

$R$

$R^{2}$

Range is same for complimentary angles.
$T_{1} = \frac{2 u s in θ}{g}$ and $T_{2} = \frac{2 u s in ( 90 - θ )}{g}$
and $R = \frac{u ^{2} s in 2 θ}{g}$
$∴ T_{1} T_{2} = \frac{2 u s in θ}{g} \times \frac{2 u cos θ}{g} = \frac{2 R}{g}$

Q. A projectile can have the same range ′R′ for two angles of projection. If T1​ and T2​ be the time of flights in the two cases, then the product of the two time of flights is directly proportional to

1/R2

1/R

R

R2

Range is same for complimentary angles. T1​=g2usinθ​ and T2​=g2usin(90−θ)​ and R=gu2sin2θ​ ∴T1​T2​=g2usinθ​×g2ucosθ​=g2R​

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

$1/ R^{2}$

$1/ R$

$R$

$R^{2}$

Range is same for complimentary angles.
$T_{1} = \frac{2 u s in θ}{g}$ and $T_{2} = \frac{2 u s in ( 90 - θ )}{g}$
and $R = \frac{u ^{2} s in 2 θ}{g}$
$∴ T_{1} T_{2} = \frac{2 u s in θ}{g} \times \frac{2 u cos θ}{g} = \frac{2 R}{g}$