Final answer:
The question involves solving a projectile motion problem by decomposing the initial velocity into horizontal and vertical components to determine the time the baseball is in the air and the maximum height it achieves.
Step-by-step explanation:
The question deals with projectile motion, which involves analyzing the vertical and horizontal components separately. First, we find the horizontal (vx) and vertical (vy) components of the initial velocity:
- vx = v * cos(θ) = 20.7 m/s * cos(21°)
- vy = v * sin(θ) = 20.7 m/s * sin(21°)
For part a, we use the horizontal distance and the horizontal component of the velocity to find the time in the air since horizontal velocity is constant:
- Horizontal Distance = vx * time
- 29.2269 m = (20.7 m/s * cos(21°)) * time
- time = 29.2269 m / (20.7 m/s * cos(21°))
For part b, we use the vertical component of the initial velocity and the time to find the maximum height using the kinematic equation for vertical motion under constant acceleration due to gravity (h = vy * t - 0.5 * g * t^2, where g is the acceleration due to gravity):
- Height = vy * (time/2) - 0.5 * g * (time/2)^2
- Height = (20.7 m/s * sin(21°)) * (time/2) - 0.5 * 9.81 m/s^2 * (time/2)^2
The time at which the ball reaches its maximum height is half of the total time in the air because of the symmetry of the projectile's path.