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.
Expert's answer

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),


"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}".

Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!


No comments. Be the first!

Leave a comment

New on Blog