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 horizontal distance of 10 m from the point of projection. Find the initial speed of projection. `(g = 10m//s^(2))`

Explanation

`theta = 45^(@), g = 10m//s^(2)`
Horizontal distance (x) = 10m
Vertical distance (y) = 7.5 m
`y = (tan theta)x - ((g)/(2u^(2) cos^(2) theta))x^(2)`
`7.5 = tan 45^(@) xx 10 - (10)/(2u^(2) cos^(2) 45^(@)) xx 10^(2)`
`7.5 = 10 - (1000)/(2.5)`
`(1000)/(u^(2)) = 2.5 rArr u^(2) = (1000)/(2.5)`
`u^(2) = 400`
`u = sqrt(400)`
`u = 20m//s`