Answer:
To find the height of the cannonball above the ground 4.30 seconds after it is fired horizontally, you can use the following kinematic equation:
\[h = \frac{1}{2}gt^2\]
Where:
- \(h\) is the height above the ground.
- \(g\) is the acceleration due to gravity, which is approximately \(9.81 \, \text{m/s}^2\) on Earth.
- \(t\) is the time in seconds.
Now, plug in the values:
\[h = \frac{1}{2} \cdot (9.81 \, \text{m/s}^2) \cdot (4.30 \, \text{s})^2\]
\[h = \frac{1}{2} \cdot 9.81 \, \text{m/s}^2 \cdot 18.49 \, \text{s}^2\]
Now, calculate:
\[h = 0.5 \cdot 9.81 \cdot 18.49 \, \text{m}\]
\[h \approx 90.72 \, \text{meters}\]
So, the cannonball is approximately 90.72 meters above the ground 4.30 seconds after it is fired horizontally.
Step-by-step explanation: