InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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» Correct Answer - (a) `(40)/(sqrt(3)) m s^-1` ; (b) `(40)/(sqrt(3)) m ; c. 40 m`. (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`. |
|