A particle is projected from the ground with some initial velocity making an angle of `45^(@)` with the horizontal. It reaches a height of 7.5 m above the ground while it travels a horzontal distance of 10 m from the point of projection . Find the initial speed of projection ( g = 10 `m//s^(2)`) .

Explanation

Angle of projection ` = 45^(@)`
vertical height , `h_(y) = 7 . 5 m `
Horizontal distance , `h_(x) = 10 m `
From fig. tan `theta = (y)/(x) = (7.5)/(10) = 3 //4 , cos theta = (4)/(5)`
`s_(x) = u * cos theta * t rArr 10 = 4 * (1)/(sqrt(2))*t`
`rArr t = (10sqrt(2))/(4) " " rarr (1)`
`s_(y) = 7 . 5 = u sin theta t - (1)/(2) gt^(2)`
` 7 .5 = (u)/(sqrt(2))(10sqrt(2))/(u) - (1)/(2) xx(10 xx 10xx 10xx 2)/(u^(2))`
` 7 . 5 = 10 - (1000)/(u^(2)) rArr - 2 . 5 = (-1000)/(u^(2))`
`rArr u^(2) = (1000)/(2 . 5 ) = 400 `
`u^(2) = 400 rArr u = 20 m//s`