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The direction of motion of a projectile ...

The direction of motion of a projectile at a certain instant is inclined at an angle `60^(@)` to the horizontal. After `sqrt(3)` seconds it is inclined an angle `30^(@)`. If the horizontal compound of velocity of projection is `3n` (in `m//sec)` then value of n is : `(take g = 10 m//s^(2))`

Text Solution

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The correct Answer is:
5
The direction ………………

Now `a_(x) = 0`
`:. u cos alpha = v cos beta`. ………. (1)
Now for motion along y axis
`a_(y) = -g`
`:. u sin alpha - "gt" = v sin beta` ………. (2)
`V = (U cos alpha)/(cos beta)`in
we have.
`u sin alpha - "gt" = (u cos alpha)/(cos beta)sdin beta`
or `u sin alpha - u cos alpha tan beta = "gt"`
`u {sin alpha - cos alpha tan beta} = "gt"`
`u = ("gt")/((sin alpha - cos alpha tan beta))`
horizontal component of velocity `= u cos alpha`.
`= ("gt" cos alpha)/((sin alpha - cos alpha tan beta)) = ("gt")/((tan alpha - tan beta))`
`= (10xxsqrt(3))/((sqrt(3)-(1)/(sqrt(3))))= 15 m//sec`.
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