Question

A particle is projected with velocity
20√3 m/s at angle of 60° with
horizontal. The speed of particle when
its velocity becomes perpendicular to
velocity of projection, is (Neglect air
friction)​

Answer 1

Answer:

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Answer 2

Answer:

$20 \ m/s$

Step-by-step explanation:

• In this question, we have been given that the particle is projected with a velocity of $20\sqrt{3} \ m/s$ at an angle of $60$° with horizontal.
• Hence, the vertical component will have, θ = $90$° - $60$° = $30$°
• The horizontal component will be taken as constant.

So, $20\sqrt{3} \ cos60$° = $vcos30$°

$20\sqrt{3} \times \frac{1}{2} = v \times \frac{\sqrt{3} }{2}$

$10\sqrt{3} = v \times\frac{\sqrt{3} }{2}$

$v = 20 \times \frac{\sqrt{3} }{\sqrt{3}}$

$v = 20 \ m/s$

Hence, the speed of the particle when its velocity becomes perpendicular to the velocity of projection is $20 \ m/s$.

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