Ablue ball is thrown upward with an initial speed of 20.6 m/s, from a height of 0.7 meters above the ground. 2.5 seconds after the blue ball is thrown, a red ball is thrown down with an initial speed of 6 m/s from a height of 23.9 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/s^2. how long after the blue ball is thrown are the two balls in the air at the same height?

The two balls are at the same height 3.6 s after the blue ball is thrown.

Explanation:

The height and velocity of the balls at any given time can be calculated using the following equations:

y = y0 + v0 · t + 1/2 · g ·t²

v = v0 + g · t

Where:

y = height at time "t".

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

v = velocity at time "t".

First, let´s calculate the height of the blue ball when the red ball is thrown (The origin of the frame of reference is located on the ground):

y = y0 + v0 · t + 1/2 · g ·t²

y = 0.7 m + 20.6 m/s · 2.5 s - 1/2 · 9.81 m/s² · (2.5 s)²

y = 21.5 m

Now let´s calculate the velocity of the blue ball at t = 2.5 s.

v = v0 + g · t

v = 20.6 m/s - 9.81 m/s² · 2.5 s

v = -3.9 m/s

Now, let´s calculate at which time after the red ball is thrown both balls have the same height. Notice that when the red ball is thrown, the blue ball is at a height of 21.5 m and its velocity is -3.9 m/s, so its initial height will be 21.6 m and the initial velocity will be -3.9 m/s:

height blue ball = height red ball

y0 + v0 · t + 1/2 · g ·t² = y0 + v0 · t + 1/2 · g ·t²

21.5 m - 3.9 m/s · t - 1/2 · 9.81 m/s² · t² = 23.9 m - 6 m/s · t - 1/2 · 9.81 m/s² · t²

21.5 m - 3.9 m/s · t -4.9 m/s² · t² = 23.9 m - 6 m/s · t - 4.9 m/s² · t²

6 m/s · t - 3.9 m/s · t = 23.9 m - 21.5 m

2.1 m/s · t = 2.4 m

t = 2.4 m/ 2.1 m/s

t = 1.1 s

The two balls are at the same height (2.5 s + 1.1 s) 3.6 s after the blue ball is thrown.

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