A ball of mass `m` is thrown at an angle of `45^(@)` to the horizontal,, from the top of a `65 m` high tower `AB` as shown in Fig. at `t = 0`. Another identical ball is thrown with velocity `20 m//s` horizontally towards `AB` from the top of a `30 m` high tower `CD 1 s` after the projection of the first ball. Both the balls move the same vertical plane. If they collide in mid-air,

the distance `AC` is

Options

A. `20m`
B. `30m`
C. `40m`
D. `20m`

Answer

C

Explanation

When two balls collide, their height from ground is same. Let it be `h`. Then vertical displacement of the first and second balls is `(65 - h) m` and `(30 - h) m`, respectively. Let the time of flight of the first ball up to the instant be `n` seconds. Then the time of flight that of second ball is `(n - 1)` seconds. Considering vertically downward component of motion of first ball,
`u=10sqrt(2)sin45^(@)=10m//s`
`a=g=10ms^(-1), s=y_(1)=(65-h)t=ns`
`s=ut+1/2at^(2)`
`(65-h)=10n+5n^(2)`………i
From eqn i and ii `n=2s, h=25m`
Displacement `AC=` Horizontal distance moved by first ball up to the instant of collision `+` that moved by the second ball up to the same instant
`:.AC = (10sqrt(2)cos45^(@))n+20(n-1)=40m`
Just before collision, vertically downward component of velocity of first ball is
`v_(y1)=(10sqrt(2)sin45^(@))+gn=30m//s`
and that of second ball is `v_(y2)=g(n-1)=10 m//s`
According to the law of conservation of momentum, vertically downward component `v_(y)`, of velocity of combined body is given by
`mv_(y1)+mv_(y2)=(m+m)v_(y)` or `v_(y)=20 ms^(-1)`
Similarly, horizontally leftward component `y`, of combined body (just after collision) is given by
`mxx20-mxx10sqrt(2)cos45^(@)=(m+m)v_(x)`
`implies v_(x)=5m//s`