A particle is projected with a velocity of 30 m - s at two angles of 30- and 60- respectively- Then the ratio of their times of flight isA- -3- 1B- 1- -3C- 1- 2D- 2- 1

The correct option is B 1 : √(3)Time of flight, T nbsp; = 2u sinθg T_1 = 2u sin 30^∘g nbsp; and nbsp; T_2 = 2u sin 60^∘g Hence T_1T_2 = sin 30^∘sin 60^ ...

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Standard XII

Physics

Projectile Time, Height and Range

A particle is...

Question

A particle is projected with a velocity of $30 m / s$ at two angles of $30^{\circ}$ and $60^{\circ}$ respectively. Then the ratio of their times of flight is

\sqrt{3} : 1

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1 : \sqrt{3}

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1 : 2

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2 : 1

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Solution

The correct option is B $1 : \sqrt{3}$
$Time of flight, T = \frac{2 u sin θ}{g} T_{1} = \frac{2 u sin 30^{\circ}}{g} and T_{2} = \frac{2 u sin 60^{\circ}}{g}$
Hence $\frac{T_{1}}{T_{2}} = \frac{sin 30^{\circ}}{sin 60^{\circ}} \frac{T_{1}}{T_{2}} = \frac{1}{\sqrt{3}}$

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A particle is projected with a velocity of 30 m/s at two angles of 30∘and 60∘ respectively. Then the ratio of their times of flight is

The correct option is B 1:√3 Time of flight, T=2usinθgT1=2usin30∘g and T2=2usin60∘g Hence T1T2=sin30∘sin60∘T1T2=1√3

A particle is projected with a velocity of $30 m / s$ at two angles of $30^{\circ}$ and $60^{\circ}$ respectively. Then the ratio of their times of flight is

The correct option is B $1 : \sqrt{3}$
$Time of flight, T = \frac{2 u sin θ}{g} T_{1} = \frac{2 u sin 30^{\circ}}{g} and T_{2} = \frac{2 u sin 60^{\circ}}{g}$
Hence $\frac{T_{1}}{T_{2}} = \frac{sin 30^{\circ}}{sin 60^{\circ}} \frac{T_{1}}{T_{2}} = \frac{1}{\sqrt{3}}$