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

3

Explanation

The horizontal 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 rArr t_1 = d//u`
` 1 = ut_1 - g/2t_(1)^(2) = u d/u - g/2 (d^2)/u^2 rArr 1 = d- g/2 d^2/u^2`
`rArr 1 = d - gd^2/(4g) rArr 4 = 4d - d^2`
`rArr d^2 - 4d + 4 = 0 rArr d = 2m `
At C : 3d ` = ut_2 rArr t_2 = (3d)/u`
`-l = ut_2 = 1/2 "gt"_(2)^(2)`
`= u* (3d)/4 - g/2 (9d^2)/(4^2) = 3d - (9gd^2)/(4g)`
`= 3d - (9d^2)/4 = 3 xx 2 - 9/4 xx 4 = 6-9 = -3`
`rArr l = 3m` .