A blue ball is thrown upward with an initial speed of 23.5 m/s, from a height of 0.7 meters above the ground. 2.9 seconds after the blue ball is thrown.

Question

A blue ball is thrown upward with an initial speed of 23.5 m/s, from a height of 0.7 meters above the ground. 2.9 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 12 m/s from a height of 31.2 meters above the ground. The force of gravity due to the earth results in the balls each having a constant downward acceleration of 9.81 m/s².

1) What is the speed of the blue ball when it reaches its maximum height?

2)How long does it take the blue ball to reach its maximum height?

3)What is the maximum height the blue ball reaches?

4)What is the height of the red ball 3.77 seconds after the blue ball is thrown? 

5)How long after the blue ball is thrown are the two balls in the air at the same height?

can you please add some notes to it as well while calculating

Answer 1

Let take vertical axis y and origin o on the grounnd. Then for blue: y₁ = 0.7 + 23.5t - 4.905t², v₁ = 23.5 - 9.81t

For red ball: y₂ = 31.2 - 12(t - 2.9) - 4.905(t - 2.9))², v₂ = - 12 - 9.81(t - 2.9)

1) At maximum height v₁ = 0

2) v₁ = 23.5 - 9.81t = 0, t = 23.5/9.81 = 2.396 s

3) To get maximum height we calculate y₁ at t = 2.396 s; y_1max= 0.7 + 23.5·2.396 - 4.901·2.396² = 28.847 m

4) y₂ at = 3.77 s, y₂ = 31.2 - 12(3.77 - 2.9) - 4.905(3.77 - 2.9)² = 17.047 m

5) y₁ = y₂; 0.7 + 23.5t - 4.905t² = 31.2 - 12(t - 2.9) - 4.905(t - 2.9))²,

0.7 - 31.2 - 12·2.9 + 4.905·2.9² = - 12t - 23.5t + 9.81·2.9t, t = 3.411 s at height 23.79 m