A small body is released from point A of smooth parabolic path `y=x^(2)`, where y is vertical axis and x is horizontal axis at ground, as shown. The body leaves the surface from point B. If `g = 10ms^(-2)` then what is the value of d (in m)?

Answer

2

Explanation

`H_(1) = (sqrt(2))^(2) =2 m`
`H_(2) = (1)^(2) = 1m`
`:. V = sqrt(2xx g xx (H_(2) - H_(1)))`
`v = sqrt(20) m s^(-1)`
`tan theta = (dy)/( dx) |_(x=1) = 2 x|_(x=1) =2 `
`- H_(2) = d tan theta - (g d^(2))/(2 u^(2) cos^(2) theta)`
`-1 = 2d - (10d^(2))/(2 (sqrt(20))^(2)(1//sqrt(3))^(2))`
`(5)/(4) d^(2) - 2 d -1 =0`
On solving we get d =2 m