An object is projected with a velocity of `20 ms^(-1)` making an angle of `45^(@)` with horizontal. The equation for the trajectory is `h = Ax -Bx^2`, where h is height, x is horizontal distance A and B are constants. The ratio A : B is `(g = 10 ms^-2)`

Options

A. `1:5`
B. `5:1`
C. `1:40`
D. `40:1`

Explanation

Time of ascent,`t=u/g=6`
`u=6g`
Distance travelled, `h=ut-1/2g t^2`
In `1^(st)` second,
`h_1=[6gxx1]-[1/2xxgxxl^2]=(11g)/2`
Distance travelled in `7^(th)` second is equal to the distance travelled in `1^(st)` second,
`h_2=[0xx7]-[1/2xxgxxI^2]=(-g)/2`
`h_1/h_2=((11g)/(2)) (-g/2)=(-11)/1`
`impliesh_1:h_2=-11:1`
`:.|h_1:h_2|=11:1.`