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The path followed by a body projected al...

The path followed by a body projected along y axis is given by `y=sqrt(3)x-(1//2)x^(2)` . If `g=10ms^(2)`, then the initial velocity of projectile will be – (x and y are in m)

A

`3 sqrt(10) m//s`

B

`2 sqrt(10)m//s`

C

`10sqrt(3)m//s`

D

`10 sqrt(2)m//s`

Text Solution

Verified by Experts

The correct Answer is:
B
Given, that `y=sqrt(3)x-(1//2)x^2`
The above equation is similar to equation of trajectory of the projectiles
`y = "tan" theta x- 1//2 (g)/(u^(2) cos^(2) theta) x^(2)`
Comparing (A) & (B) we get
`tan theta = sqrt(3) rArr theta = 60^(@)`
and `1//2 =(1//2)(g)/(u^(2)cos^(2)theta)`
`rArr" "u^(2)cos^(2) theta =g`
`rArr " " u^(2) cos^(2) 60 =10`
`rArr u^(2)(1//4)=10`
`rArr " " u^2 = 40 `
`rArr " " u=2 sqrt(10)m//s`
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