Sure, I'd be happy to help with that!
a) To find the highest point of the projectile's trajectory, you can use the formula for the maximum height of a projectile: \( H = \frac{v^2 \cdot \sin^2(\theta)}{2g} \), where \( H \) is the maximum height, \( v \) is the initial velocity (20 m/s), \( \theta \) is the launch angle (30°), and \( g \) is the acceleration due to gravity (9.8 m/s²). By plugging in these values, you can calculate the maximum height as follows:
\( H = \frac{20^2 \cdot \sin^2(30°)}{2 \cdot 9.8} \)
\( H = \frac{400 \cdot (\frac{1}{2})^2}{19.6} \)
\( H = \frac{400 \cdot \frac{1}{4}}{19.6} \)
\( H = \frac{100}{19.6} \)
\( H \approx 5.10 \, \text{meters} \)
Therefore, the highest point of the projectile's trajectory is approximately 5.10 meters.
b) To determine the horizontal range of the projectile, you can use the formula for the range of a projectile: \( R = \frac{v^2 \cdot \sin(2\theta)}{g} \). Using the same initial values, you can calculate the horizontal range as follows:
\( R = \frac{20^2 \cdot \sin(2 \cdot 30°)}{9.8} \)
\( R = \frac{400 \cdot \sin(60°)}{9.8} \)
\( R = \frac{400 \cdot \frac{\sqrt{3}}{2}}{9.8} \)
\( R = \frac{200 \sqrt{3}}{9.8} \)
\( R \approx 35.3 \, \text{meters} \)
Therefore, the horizontal range of the projectile is approximately 35.3 meters.
c) To calculate the time the projectile is in the air, you can use the formula for the total time of flight: \( T = \frac{2v \cdot \sin(\theta)}{g} \). Plugging in the given values:
\( T = \frac{2 \cdot 20 \cdot \sin(30°)}{9.8} \)
\( T = \frac{40 \cdot \frac{1}{2}}{9.8} \)
\( T = \frac{20}{9.8} \)
\( T \approx 2.04 \, \text{seconds} \)
Therefore, the projectile is in the air for approximately 2.04 seconds.
