A projectile is launched at time t = 0 from point A which is at height 1 m above the floor with speed `v ms^(-1)` and at and angle `theta = 45^@` with the floor. It passes through a hoop at B which is 1 m above A and B is the highest point of the trajectory. The horizontal distance between A and B is d meters. The projectile then falls into a basket, hitting the floor at C a horizontal distance 3 d meters from A. Find l (in m).

Answer

C

Explanation

The horizontally and vertical components of the velocity are the same, let it be `u=v cos 45^(@)`
From `A` to `B` :`1=u^(2)/(2g) rArr u^(2)=2g`
At `B:d=ut_(1)rArrt_(1)d/u,1=ut_(1)g/2t_(1)^(2)=ud/u-g/2d^(2)/u^(2)`
`rArr 1=d-g/2d^(2)/u^(2) rArr 1=d-(gd^(2))/(4g)`
`rArr 4=4d=d^(2) rArr d^(2)-4d+4=0 rArr d=2m`
`3d=ut_(2) rArr 1t_(2)=(3d)/u`
`-1=ut_(2)-1/2"gt"_(2)^(2)=-u.(3d)/4-g/2(9d^(2))/4^(2)=3d-(9gd^(2))/(4g)`
`=3d-(9d^(2))/(4g)=3xx2-9/4xx4=6-9=-3`
`rArr l=3m`