A projectile is projected at 10 ms^(-1) ...

A projectile is projected at `10 ms^(-1)` by making at an angle `60^(@)` to the horizontal. After some time its velocity makes an angle of `30^(@)` to the horizontal. Its speed at this instant is

`10/sqrt(3)`

`10sqrt(3)`

`5/sqrt(3)`

5sqrt(3)`

Verified by Experts

The correct Answer is:

During projectile motion horizontal component of force remains the same.
`:. Vcostheta=vcosalpha`
`10xxcos60^@=vcos30^@`
`:.v=(10xx1/2)/(sqrt(3)//2)`

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