Final answer:
The ball is in the air for approximately 1.5 seconds. The initial horizontal component of the velocity is 13.0 m/s. The vertical component of the velocity just before the ball hits the ground is approximately -14.3 m/s.
Step-by-step explanation:
The ball will follow a parabolic path due to the combination of its horizontal and vertical motion. To answer the questions (a) to (d), we can break down the ball's motion into separate horizontal and vertical components.
(a) To calculate the time the ball is in the air, we need to determine the time it takes for the ball to reach the ground vertically. We can use the equation:
h = (1/2)gt^2
where h is the initial height (23.2 m) and g is the acceleration due to gravity (9.8 m/s^2). Plugging in the values and solving for t, we find that the ball is in the air for approximately 1.5 seconds.
(b) The initial horizontal component of the velocity remains constant throughout the motion. Since the ball is thrown horizontally, the horizontal component of the velocity is equal to its initial velocity. Therefore, the initial horizontal component of the velocity is 13.0 m/s.
(c) To find the vertical component of the velocity just before the ball hits the ground, we can use the equation:
v = u + gt
where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity, and t is the time. Plugging in the values, we find that the vertical component of the velocity just before the ball hits the ground is approximately -14.3 m/s (negative because it is directed downwards).
(d) To find the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground, we can use the Pythagorean theorem:
velocity = sqrt((horizontal velocity)^2 + (vertical velocity)^2)
Plugging in the values, we find that the velocity of the ball just before it hits the ground is approximately 18.1 m/s.