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.

Calculate the distance d between the towers.

Answer

`10sqrt(3)m`

Explanation

Let collision occurs at time t
For horizontal motion of m ,
The displacement of A, PM = 10 t ... (i)
For vertical motion of m
(Taking + y in vertically downward direction)
`u_y = 0, s_y = y and a_y = g`
Using `s_y = u_yt + 1/2 a_yt^2 = 1/2 gt^2" ".....(ii)`
and `V_y = u + at = gt" " ... (iii)`
For mass 2m thrown from C
`u_x = 10 cos 60^@ = 10 xx 1/2 = 5m//s , v_y = 10 sin 60^@ = 10 xx (sqrt3)/2 = 5 sqrt(3) m//s`
For horizontal motion `QM = u_xt = 5t" " . . .(iv)`
For vertical motion
`V_y = 5sqrt(3) + g t" ".....(v)`
`s_y = u_yt + 1/2 a_yt^2 = 5sqrt(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 implies t = 2/(sqrt3) "sec"`
`:. BD = PM + MQ = 10t + 5t = 15t = 15 xx 2/(sqrt3) , " " = 10sqrt(3) = 17.32 m`.