Question

A ball is thrown vertically upward from the 40-ft level in an elevator shaft with an initial velocity (straight up) of 50 ft/sec. At the same instant, far below, an open-platform elevator passes the 10-ft level in the elevator shaft, moving upward with a constant velocity of 5 ft/sec. Determine (a) when and where the ball will hit the elevator, (b) the relative velocity of the ball with respect to the elevator when the ball hits the elevator.

Answer 1

here

g = 9.8 m/s^2 = 32.15 ft/s^2

motion of ball

vb = v0 + a*t

vb = 50 - 32.15 * t

then

yb = y0 + v0*t + 0.5 * a*t^2

yb = 40 + 50 * t - 16 *t^2

then the motion of elevator

ve = 5ft/sec

ye = 10 + 5 * t

then

ye = yb

10 + 5 *t = 40 + 50 *t - 16 *t^2

16 *t^2 - 45 *t -30 = 0

by solving this quadratic equation we get

t = 3.37 sec and -0.556 sec

we take the positive value of the t

so

ye = 10 + 5 * 3.37 = 26.85 ft

elevation from ground is 26.85 ft

the relative velocity with respect to the elevator is

vb - ve = (50 - 32.15 *t ) - 5  

vb - ve = (50 - 32.15 * 3.37) - 5

vb - ve = -53.34 ft/sec

the negative sign means that elevator is moving downward