A particle is projected from ground with speed $80 m / s$ at an angle $30^{o}$ with horizontal from ground.The magnitude of average velocity of the particle in time interval $t = 2 s$ to $t = 6 s$ is [Take $g = 10 m / s^{2}$ ]

40 \sqrt{2} m / s

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40 m / s

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Z e r o

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40 \sqrt{3} m / s

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Solution

The correct option is D $40 \sqrt{3} m / s$
Time of flight = $\frac{2 u sin θ}{g}$
= $\frac{2 \times 80 \times sin 30^{o}}{10} = 8$
As particle during journey at $t = 2 s e c$ and at $t = 6 s e c$ will be at same height and hence vertical displacement is zero.
Horizontal displacement = $(u cos θ) \times (6 - 2)$
= $80 \times \frac{\sqrt{3}}{2} \times 4$
= $160 \sqrt{3} m$
$∴$ Average velocity = $\frac{160 \sqrt{3}}{4} = 40 \sqrt{3} m / s e c$

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A particle is projected from ground with speed 80m/s at an angle 30o with horizontal from ground.The magnitude of average velocity of the particle in time interval t=2s to t=6s is [Take g=10m/s2]

A particle is projected from ground with speed $80 m / s$ at an angle $30^{o}$ with horizontal from ground.The magnitude of average velocity of the particle in time interval $t = 2 s$ to $t = 6 s$ is [Take $g = 10 m / s^{2}$ ]