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Equation of trajector of ground to groun...

Equation of trajector of ground to ground projectile is `y=2x-9x^(2)`. Then the angle of projection with horizontal and speed of projection is : `(g=10m//s^(2))`

A

` theta_(0) = cos^(-1) ((1)/( sqrt(5))) and v_(0) = (5)/(3) ms^(-1)`

B

` theta_(0) = sin ^(-1) ((1)/( sqrt(5))) and v_(0) = (5)/( 3) ms^(-1)`

C

` theta_(0) = cos ^(-1) ((2)/( sqrt(5))) and v_(0) = (3)/( 5) ms^(-1)`

D

` theta_(0) = sin ^(-1) ((2)/( sqrt(5))) and v_(0) = (3)/( 5) ms^(-1)`

Text Solution

Verified by Experts

The correct Answer is:
A
` y = 2 x - 9 x ^(2), y = x tan theta - (1)/(2) ( gx ^(2))/( u^(2) cos ^(2) theta )`
`tan theta = 2 cos theta = (1)/( sqrt(5)) rArr theta = cos ^(-1) ((1)/( sqrt(5))) `
`( g)/( 2 u^(2) cos ^(2) theta)= 9 rArr ( g (1 + tan^(2) theta))/( 2 u^(2)) = 9`
`rArr (10 (1 + 4))/(2 u^(2)) = 9`
` u^(2) = (10 xx 5)/( 2 xx 9) rArr u = (5)/( 3) m//s`
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