Answer to Question #133445 in Engineering for Isaac musonda

Question

Ball thrown vertically from 12 m level in elevator
shaft with initial velocity of 18 m/s. At same
instant, open-platform elevator passes 5 m level
moving upward at 2 m/s.
Determine (a) when and where ball hits elevator and
(b) relative velocity of ball and elevator at contact.

Answer 1

Let us write equations of motion of the ball and the elevator:

$y_b(t) = H_b + v_{0b} t - \frac{g t^2}{2}$, $v_b(t) = v_{0b} - g t$,

$y_e(t) = H_e + v_{0e} t$, $v_e(t) = v_{0e}$ (elevator moves with constant speed upwards),

where:

$H_b = 12 m, H_e = 5 m$ are initial heights of ball and elevator respectively,

$v_{0b} = 12 \frac{m}{s}, v_{0e} = 2 \frac{m}{s}$ are initial velocities of ball and elevator respectively.

a) When the ball and elevator contact, $y_b(t) = y_e(t)$, from where $H_b + v_{0b} t - \frac{g t^2}{2} = H_e + v_e t$, or, rearranging terms, $\frac{g t^2}{2} + (v_{0e} - v_{0 b})t + (H_e - H_b) = 0$.

This is a quadratic equation for time, when they contact. Solving the equation, obtain two roots $t_1 = 3.65, t_2 = -0.39$. We discard the negative solution, hence the ball hits the elevator in $t_1 = 3.65s$ at $y_e(t_1) \approx 12.3 m$.

b) The velocity of the ball at time $t_1$ is $v_b(3.65) \approx -23.81 \frac{m}{s}$ (it is negative, because the ball is moving down). Velocity of the elevator remains $2 \frac{m}{s}$, and is directed up, so relative velocity is $v_r = 23.81 \frac{m}{s} + 2 \frac{m}{s} = 25.81 \frac{m}{s}$.