Final answer:
The ball reaches a height of 24.6 m and is in the air for a total time of 2.24 s before being caught.
Step-by-step explanation:
In the vertical direction, the initial velocity is 22 m/s and we can assume that the final velocity is 0 m/s at the highest point of the ball's trajectory. Using the kinematic equation, we can determine the displacement (height) of the ball using the following formula:
Δy = v₀yt + (1/2)gt², where v₀y is the initial velocity in the vertical direction, t is the time, and g is the acceleration due to gravity (approximately 9.8 m/s²).
Substituting the given values, we have Δy = (22 m/s)(t) - (1/2)(9.8 m/s²)(t²) = 0.
Rearranging the equation and solving for t, we get t = 2.24 s.
Substituting the value of t back into the previous equation, we can find the height of the ball:
Δy = (22 m/s)(2.24 s) - (1/2)(9.8 m/s²)(2.24 s)² = 24.6 m.
In the horizontal direction, the initial velocity is 0 m/s and the final velocity is also 0 m/s because the ball returns to the same horizontal position.
Using the equation d = v₀xt, where d is the distance, v₀x is the initial velocity in the horizontal direction, and t is the time, we can determine the time it takes for the ball to return to the initial position.
Substituting the given values, we have 0 = (16 m/s)t.
Since the initial velocity is 0 m/s, the time it takes for the ball to return to the initial position is t = 0 s.