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A ball is projected with a velocity 20 s...

A ball is projected with a velocity `20 sqrt(3) ms^(-1)` at angle `60^(@)` to the horizontal. The time interval after which the velocity vector will make an angle `30^(@)` to the horizontal is (Take, `g = 10 ms^(-2))`

A

5s

B

2s

C

1s

D

3s

Text Solution

Verified by Experts

The correct Answer is:
B
`tan 30^(@) = (v_(y))/(v_(x)) = (u_(y) - g t)/(u_(x))`
`= ((20sqrt(3)sin 60^(@))-10t)/((20sqrt(3)cos 60^(@)))`
or `10 = 30 - 10 t`
`:. T = 2s`
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