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.

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),

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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