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.

Answer

A, B, C, D

Explanation

(i) 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 , t_(y) = t `
`:. y = 1/2 gt^(2)` ....(ii)
` v_(y) = u_(y) + a_(y)t = gt` ...(iii)
For mass `2 m` thrown from `C`
For horizontal motion ` QM = [ 10 cos 60(@)] t `
` QM = 5 t` ....(iv)
For vertical motion `v_(y) = 10 sin 60(@) = 5 sqrt (3), a_(y) = g ` ,brgt ` s_(y) = y + 10, t^(y) = t `
Now , `v_(y) = 5 sqrt(3) + gt ` ..... (v)
and `(s_(y)) = u _(y)t + (1)/(2) a_(y) t^(2)`
`rArr y + 10 = 5 sqrt(3)t + (1)/(2) g t^(2) ` ...(vi)
From (ii) and (vi)
`:. (1)/(2) g t^(2) ` +10 = 5sqrt(3)t + 1/2 g t^(2) rArr t = (2)/sqrt(3) sec `
`:. BD = PM + MQ = 10 t + 5 t = 15 t = 15xx (2)/(sqrt(3))`
` = 10sqrt(3) = 17.32 m`
(ii) Applying conservation of the linear momentum (during collision of the mass at ` M`) in the horizontal direction
` 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 = 10t = 10 xx (2)/(sqrt(3)) = 11.547 m `