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A ball is thrown vertically upward from the 12m level with u=18m/s .At the same instant an open platform elevator passes the 5m level ;moving upward with a constant velocity of 2m/s . Determine:-1. when and where the ball will meet the elevator 2.the relative velocity of the ball with respect to the elevator when ball hits the elevator .Sir please send me a brief and full understandable answer

A ball is thrown vertically upward from the 12m level with u=18m/s .At the same instant an open platform elevator passes the 5m level ;moving upward with a constant velocity of 2m/s . Determine:-1. when and where the ball will meet the elevator 2.the relative velocity of the ball with respect to the elevator when ball hits the elevator .Sir please send me a brief and full understandable answer

Grade:12

1 Answers

Vikas TU
14149 Points
6 years ago
yb = 12 + 18t - 4.9t^2            [Equation for ball]
yp = 5 + 2t                             [Equation for platform elevator]
The Ball will meet the elevator when both yb and yp are equal.
5 + 2t = 12 + 18t - 4.9 t^2 
4.9 t^2 - 16 t - 7 = 0 
Solving the above quadratic equation using  one root will be negative, so is not relevant. 
The other root is 3.65
Hence the ball will meet the elevator at 3.65 sec
and by solving y = 5 + 2t the distance where elevator meets ball can be found
Substituting t = 3.65
y= 12.3 m

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