A particle is projected with a velocity 10 m//s at an angle `37^(@)` to the horizontal. Find the location at which the particle is at a height 1 m from point of projection.

Answer

1.6 m, 8 m.

Explanation

`y=x tan theta - (gx^(2))/(2u^(2) cos^(2) theta)`
for `y=1, theta=37^(@), u=10 m//s`
`1=3/4x - (10x^(2))/(2xx100xx((16),(25)))`
`1=3/4x -5/64 x^(2)`
`5x^(2)-48x+64 =0`
`5x^(2)-40x-8x+64=0`
`x=8m, 1.6 m`