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Two seconds after projection, a projecti...

Two seconds after projection, a projectile is travelling in a direction inclined at `30^@` to the horizontal. After one more second, it is travelling horizontally. Find the magnitude and direction of its velocity.

Text Solution

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Let us take u as the magnitude of initial velocity and 0 be the projection angle. It is given that at t = 3.0 s, stone is at maximum height. Thus we have half of time of flight 3.0s i.e
`(u sin 0)/(g)=3 rarr u sin 0=3g" ….(i)"`
If we take v the magnitude of velocity of the stone at t = 2.0 s when it is making an angle `30^(@)` with the horizontal, we have `v cos 30^(@) = u cos 0 and v sin 30^(@) = u sin 0 – g(2)`
or `v ((1)/(2))=30-20=10" or "v=20 m//s`
Now from horizontal component `v ((sqrt(3))/(2)) =u cos 0`
or `u cos 0 =10 sqrt(3)" ....(ii)"`
Squaring and adding (1) and (2), we have `u =20 sqrt(3) m//s`
Dividing we get:
`tan 0 =sqrt(3)" or "0=60^(@)`
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