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A particle is projected at an angle of `60^(@)` above the horizontal with a speed of `10m//s`. After some time the direction of its velocity makes an angle of `30^(@)` above the horizontal. The speed of the particle at this instant is s

A

`5/sqrt3 m//s`

B

`5sqrt3` m/s

C

5 m/s

D

`10/sqrt3` m/s

Text Solution

Verified by Experts

The correct Answer is:
D

Let v be the velocity of projectile at this instant , Horizontal component of velocity remains unchanged .
Therefore , `v cos 30^(@) = 10 cos 60^@`
or `vsqrt3/2=10/2" or " v =10/sqrt3m//s`
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