Answer:
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.
