Two towers AB and CD are situated at 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 fromt the top of CD at an angle of `60^(@)` to the horizontal towards AB with the same magnitude of initial velocity as that oft he first object. The two objects move in the same vertical plane, collide in mid air and stick to each other (i) calculate the distance between the towers and (ii) find the position where the objects hit the ground.

Explanation

Method 1 : (a) Let `t` be the time taken for collision. For mass `m` thrown horizontally from `A` for horizontal motion,
`PM = 10 t` …(i)
For vertical motion, `u_y = 0, s_y = y, a_y = g`
`S_y = u_y t + (1)/(2) a_y t^2 rArr y = (1)/(2) "gt"^2`...(ii)
`v_y = u_y + a_y t = "gt"` ...(iii)
From mass `2 m` thrown from `C`, for horizontal motion
`QM = [10 cos 60^@]t rArr QM = 5 t` ...(iv)
For vertical motion : `u_y = 10 sin 60^@ = 5 sqrt(3))`
`rArr v_y = 5 sqrt(3)) + "gt"` ...(v)
`a_y = g. S_y = y + 10, S = ut + (1)/(2) at^2`
`rArr y + 10 = s sqrt(3) t + (1)/(2) "gt"^2` ...(vi)
From (ii) and (vi), `(1)/(2) "gt"^2 + 10 = s sqrt(3) t + (1)/(2) "gt"^2 rArr t = (2)/(sqrt(3) s`
`BD = PM + MQ`
=`10 t + 5t`
=`15 t = 15 xx (2)/(sqrt(3)`
=`10 sqrt(3) = 17.32 m`
(b) Applying the conservation of linear momentum (during collision of the masses at M) in the horizontal direction, we have
`m xx 10 - 2m 10 cos 60^@`
=`3 m xx v_x rArr 10 m - 10 m = 3m xx v_x rArr v_x = 0`
Since the horizontal momentum comes out to be zero, the combination of masses will drop vertically downwards
and fall at `E,BE = PM = 10 t = 10 xx (3)/(sqrt(3) = 11.547 m`
Method 2 : Acceleration of `A and C` both is `9.8 m s^-2` downward.
Therefore, the 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 `vec v_(C A) = vec v_C - vec v_A` = relative velocity of `C w.r.t. A` should be along `CA`
`vec v_A = 10 hat i`
`vec v_B = -5 hat i- 5 sqrt(3) hat j`
`vec v_(B A) = - 5 hat i - 5 sqrt(3) hat j - 10 hat i :. vec v_(B A) = -15 hat i - 5 sqrt(3) hat j`
`:. tan 60^@ = (d)/(10) or d = 10 sqrt(3) m`.
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