To determine how high above the ground the coin is when its speed is 60.0 m/s, we can use the principles of kinematics. We'll assume the coin is dropped from rest (initial speed is zero) and that it is falling freely under the influence of gravity.
We can use the following kinematic equation, which relates the final velocity \( v \), initial velocity \( u \), acceleration \( a \), and displacement \( s \):
\[ v^2 = u^2 + 2as \]
where:
- \( v = 60.0 , \text{m/s} \) (final speed of the coin),
- \( u = 0 , \text{m/s} \) (initial speed, since it fell from a rest position),
- \( a = 9.81 , \text{m/s}^2 \) (acceleration due to gravity),
- \( s \) is the displacement from the point where it was released until its speed reaches 60.0 m/s.
Substituting the known values into the equation, we have:
\[ (60.0 , \text{m/s})^2 = (0 , \text{m/s})^2 + 2(9.81 , \text{m/s}^2)s \]
This simplifies to:
\[ 3600 = 19.62s \]
Solving for \( s \):
\[ s = \frac{3600}{19.62} \approx 183.4 , \text{m} \]
This displacement \( s \) is how far the coin has fallen from the point it was dropped. Now we can determine the height above the ground when it has fallen that distance.
Initially, the height of the balloon (and hence the point where the coin was dropped) was:
\[ H = 3150 , \text{m} \]
The height of the coin when it has fallen \( s \) meters is:
\[ \text{Height above the ground} = H - s = 3150 , \text{m} - 183.4 , \text{m} \approx 2966.6 , \text{m} \]
Therefore, the coin is approximately 2967 meters above the ground when its speed reaches 60.0 m/s.