Question

A ball is thrown vertically upward from the 12 m level with an initial velocity of 18 m/s. At the same instant an open platform elevator passes the 5 m level, moving upward with a constant velocity of 2 m/s. Determine (g = 9.8 $m/s^2$)
(a) when and where the ball will meet the elevator
(b) the relative velocity of the ball with respect to the elevator when the ball hits the elevator.

Answer 1

$\left(A\right)$Initial velocity of ball $v_o=+18m/s$

Initial position of ball $y_0=+12m$

Acceleration $a=-9.81m/s^2$

Substituting these values in the equation for uniformly accelerated motion we get

$v_B=v_0+at$

$v_B=18-9.81t$

$y_B=y_0+v_{0}t+\frac{1}{2}a{t}^2$

$y_B=12+18t-4.905t^2$

Now,

Initial velocity of elevator $v_E=+2m/s$

$y_E=y_0+v_{E}t$

$y_E=5+2t..........\left(1\right)$

When ball hit the elevator then

$y_E=y_B$

so,

$5+2t=12+18t-4.905t^2$

$4.90t^2-16t-7=0$

$t=-0.39or3.656$

$t=3.65sec$

Putting in (1)

$y_E=5+2\left(3.65\right)$

$y_E=12.30m$

$\left(B\right)$Relative velocity of ball wrt the elevator

$v_{B/E}=v_B-v_E=(18-9.81t)-2=16-9.81t$

When the ball hits the elevator at time $t=3.65sec$ we have

$v_{B/E}=16-9.81\left(3.65\right)$

$v_{B/E}=-19.81m/s$

The negative sign means that the ball is observed from the elevator to the moving in the negative sense(downward)