The initial velocity of a projectile of mass 1 kg is given by u- - -3-m-s- What is the time of flight-2-3-5-s-3-10-s-3-3-5-s-3-5-s-

The correct option is D √(3)/5 s Vertical component of velocity, u_y = u sin θ = √(3) m/s Now, time of flight, T=2u sin θ/g =2 ×√(3)/10=√(3)/5 s Hence, the cor ...

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Byju's Answer

Standard XII

Physics

Definition of a Projectile

The initial v...

Question

The initial velocity of a projectile of mass 1 kg is given by $u = (^i + \sqrt{3}^j) m / s$ . What is the time of flight?

$\frac{2 \sqrt{3}}{5} s$

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$\frac{\sqrt{3}}{10} s$

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$\frac{3 \sqrt{3}}{5} s$

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$\frac{\sqrt{3}}{5} s$

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Solution

The correct option is D
$\frac{\sqrt{3}}{5} s$

Vertical component of velocity, $u_{y} = u s i n θ = \sqrt{3} m / s$
Now, time of flight, $T = \frac{2 u s i n θ}{g} = \frac{2 \times \sqrt{3}}{10} = \frac{\sqrt{3}}{5} s$
Hence, the correct choice is (d).

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The initial velocity of a projectile of mass 1 kg is given by u=(^i+√3 ^j) m/s. What is the time of flight?

The correct option is D √35 s Vertical component of velocity, uy=u sin θ=√3 m/s Now, time of flight, T=2u sin θg=2×√310=√35 s Hence, the correct choice is (d).

The initial velocity of a projectile of mass 1 kg is given by $u = (^i + \sqrt{3}^j) m / s$ . What is the time of flight?

The correct option is D
$\frac{\sqrt{3}}{5} s$

Vertical component of velocity, $u_{y} = u s i n θ = \sqrt{3} m / s$
Now, time of flight, $T = \frac{2 u s i n θ}{g} = \frac{2 \times \sqrt{3}}{10} = \frac{\sqrt{3}}{5} s$
Hence, the correct choice is (d).