Question

Ball A is thrown vertically upwards with a velocity of v0. Ball B is thrown upwards from the same point with the same velocity t seconds later. Determine the elapsed time t < 2v0/g from the instant ball A is thrown to when the balls pass each other, and find the velocity of each ball at this instant.

Answer 1

Answer: ta= (gtb + 2V0)/2g

Vay=-1/2(gtb)

Vby=+1/2(gtb)

Explanation: See the attachment

Answer 2

Answer:

  t = ½ t₀ / v₀

ball ,     v = (v₀² - ½ g t²) / v₀

ball 2,   v = v₀ - g t₀ (1/2v₀  - 1)

Explanation:

We can solve this exercise with the equations of science, let's write for the first ball

                 y₁ = v₀ t - ½ g t²

To write the equation of the second ball, which was released later. To use the same time

                 t ’= t - t₀

                 y₂ = v₀ (t - t₀) - ½ g (t - t₀)²

At the point where the two balls meet the height is the same y₁ = y₂

               v₀ t - ½ g t² = v₀ (t-t₀) - ½ g (t - t₀)²

               v₀ t₀ - ½ g t² = - ½ g (t² - 2 t₀ t + t₀²)

               v₀ t₀ = g t₀ t - ½ g t₀²

               ½ g t₀² + t₀ (v₀ - g t) = 0

               t₀ (g / 2 t₀ + (v₀ - gt)) = 0

 

The solution of this equation is

              ½ g t₀ + v₀ - g t = 0

               t = ½ t₀ / v₀

This is the time for them to meet; the speeds are

Ball 1

           v = v₀ - g t

           v = v₀ - g ½ t₀ / v₀

           v = (v₀² - ½ g t²) / v₀

Ball 2

           v = v₀ - g (t - t₀)

           v = v₀ - g (½ t₀/v₀  - t₀)

           v = v₀ - g t₀ (1/2v₀  - 1)