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Grade 11Mechanics

A ball is thrown vertically upward from the 12m level with an initial velocity of 18m/s .At the same instant an open platform elevator passes the 5 m level,moving upward with a constant velocity of 2m/s.Determine -(a) when and where the ball will meet the elevator, (b)the relative velocity of the ball w.r.t.elevator when the ball hits the elevator

Profile image of Anjana Sirothiya
9 Years agoGrade 11
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1 Answer

Profile image of Vikas TU
9 Years ago
Dear Student,
.
Initial relative velocity of ball with respect to elevator = 18 -2
                                                                                           = 16m/sec
               Relative acceleration of ball with respect to elevator = -10-0
                                                                                         =-10m/sec^2
To meet the elevator ball has to travel a distance of 7m
             S = -7m (downwards)
  • S = ut+1/2 at^2
  • -7 = 16t – ½ *10 *t^2
  • -7 = 16t – 5t^2
Solve for t
  • t = (16±√396)/ 10
Neglecting  negative root
  • t = (16+√396)/10
Relative velocity at crossing
v = u+at
  •          = 16 – 10 *((16+√396)/10)
  •          = 16- 16-√396
  •          =-36 m/sec
Cheers!!
Regards,
Vikas (B. Tech. 4th year
Thapar University)