Ball `I` is thrown towards a tower at an angle of `60^@` with the horizontal with unknown speed `(u)`. At the same moment, ball `II` is released from the top of tower as shown in (Fig. 5.189). Balls collide after `2 s` and at the moment of collision, the velocity of ball `I` is horizontal. Find the
(a) speed `u`
(b) distance of point of projection of ball `I` from base of tower `(x)`
( c) height of tower
.

Answer

(a) `(40)/(sqrt(3)) m s^-1` ;
(b) `(40)/(sqrt(3)) m ; c. 40 m`.

Explanation

(a) Time of ascent should be `2 s`.
So `(u sin 60^@)/(g) = 2 rArr u = (40)/(sqrt(3)) ms^-1`
(b) `x = (u cos 60^@) t = (40)/(sqrt(3)) xx (1)/(2) xx 2 = (40)/(sqrt(3)) m`
( c) The height of tower can be found using concept of relative velocity. If the balls have to collide, the initial velocity of ball `I` should be towards ball `II`.
For this `(h)/(x) = tan 60^@ rArr h = x tan 60^@ = (40)/(sqrt(3)) sqrt(3) = 40 m`
OR : height ascended by ball `I "in" 2 s` :
`h_1 = u sin 60^@ xx 2 - (1)/(2) 10 (2)^2 = 20 m`
height decended by ball `II in 2 s` ,
`h_2 = (1)/(2) "gt"^2 = (1)/(2) 10 (2)^2 = 20 m`
Now `h = h_1 + h_2 = 20 + 20 = 40 m`.