# Answer to Question #133445 in Engineering for Isaac musonda

Question #133445
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.
1
2020-09-18T08:42:24-0400

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}".

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