Question

In the elevator shaft, a ball is thrown vertically upward with an initial velocity of 20 m/s from a height of 10 m above the ground. At the same instant, an open-platform elevator passes the 4-m level, moving upward with a constant velocity of 3 m/s. Determine (a) when and where the ball hits the elevator (b) the relative velocity of the ball with respect to the elevator when the ball hits the elevator

Answer 1

Answer:

 a) t₁= 3.256 s,  y = 17.89 m , b)   v _r = -14.9 m / s

 Explanation:

We can work this exercise with kinematic relationships

For the ball

           y₁ = y₀₁ + v₀₁ t - ½ g t²

For the platform

           y₂ = y₀₂ + v₀₂ t

Where the speed of the ball v₀₁ = 20 m / s and the initial height y₀₁ = 10 m, for the speed of the platform v₀₂ = 3 m / s and the initial height y₀₂ = 4 m

At the point where they find the position of the two is the same

             y₁ = y₂

            y₀₁ + v₀₁ t - ½ g t² = y₀₂ + v₀₎ t

            y₀₁ - y₀₂ + t (v₀₁ –v₀₂) - ½ g t² = 0

We substitute the values

           10-4 + t (20-3) - ½ 9.8 t² = 0

            6 + 17 t - 4.9 t2 = 0

            t² - 3,469 t - 1,224 = 0

We solve the second degree equation

            t = [3,469 ±√(3,469 2 - 4 1,224)] / 2

            t = [3,469 ± 3.0427] / 2

            t₁ = 3.2559 s

            t₂ = 0.2132 s

We take the longest time, the smallest time is for the ball with negative initial velocity

            t₁= 3.256 s

Let's find out how much the platform has risen in this time

             y = y₀₂ + v₀₂ t

             y = 4 + 3 3.256

             y = 17.89 m

b) let's look for the speed of the ball at this time

             v = v₀₁ - g t

             v = 20 - 9.8 3.2559

             v = -11.9 m / s

The relative speed is

             v_r = v –vo2

              v_r = -11.9 - 3

              v _r = -14.9 m / s