Two towers AB and CD are situated a distance d apart as shown in figure. AB is 20 m high and CD is 30 m high from the ground. An object of mass m is thrown from the top of AB horizontally with a velocity of 10 m/s towards CD. Simultaneously another object of mass 2 m is thrown from the top of CD at an angle `60^@` below the horizontal towards AB with the same magnitude of initial velocity as that of the first object. The two objects move in the same vertical plane, collide in mid-air and stick to each other.

Find the distance from B where the objects hit the ground.

Answer

`(20)/(sqrt3) m`

Explanation

Applying conservation of linear momentum (during collision of the masses at M) in the horizontal direction
`m xx 10 - 2m xx 5 = 3m xx V_x " " implies 10 m - 10m = 3m xx V_x implies 0`
Since the horizontal momentum comes out to be zero, the combination of masses will drop vertically downwards and fall at E.
`BE = PM = 10t = 10 xx 2/(sqrt3) = 11.54 m`
Method 2 :
Acceleration of A and C both is `10 m//s^2` downwards.
Therefore, relative acceleration between them is zero, i.e., the relative motion between them will be a straight line.
Now assuming A to be at rest, the condition of collision will be that `vecv_(CA) = vecv_C - vecv_A` = relative velocity of C w.r.t A should be along CA.
`vecv_A = 10 hati`
`vecv_C = - 5hati - 5 sqrt(3) hatj`
`vecv_(CA) = - 5 hati - 5sqrt(3)hatj - 10 hati" ":. vecv_(CA) = -15 hati - 5sqrt(3) hatj`
`implies tan theta = 15/(5sqrt3) = sqrt(3) implies theta = 60^@`
`:. tan 60^@ = d/10 " " :. d = 10sqrt(3)m`