Answer:
To find how long the ball is in the air before it reaches its highest point, you can use the following kinematic equation:
\[t = \frac{{v_i \sin(\theta)}}{g}\]
Where:
- \(t\) is the time of flight.
- \(v_i\) is the initial velocity, which is 30 m/s.
- \(\theta\) is the launch angle, which is 53 degrees.
- \(g\) is the acceleration due to gravity, which is 9.8 m/s².
First, convert the angle from degrees to radians because trigonometric functions in physics typically use radians:
\(\theta\) (in radians) = \(53^\circ\) × \(\frac{\pi}{180}\) ≈ 0.927 radians
Now, plug in the values into the equation:
\[t = \frac{{30 \, \text{m/s} \cdot \sin(0.927)}}{9.8 \, \text{m/s}^2}\]
Calculate the sine of the angle:
\[\sin(0.927) \approx 0.801\]
Now, calculate \(t\):
\[t \approx \frac{{30 \, \text{m/s} \cdot 0.801}}{9.8 \, \text{m/s}^2} \approx 2.44 \, \text{s}\]
So, the ball is in the air for approximately 2.44 seconds before it reaches its highest point.