Final answer:
The ball hits the ground at a speed of 19.1 m/s, remains in the air for 2.65 seconds, and reaches a maximum height of 8.58 meters.
Step-by-step explanation:
Given that a ball is kicked with an initial velocity of 14.0 m/s in the horizontal direction and 13.0 m/s in the vertical direction, we can determine the speed at which the ball hits the ground, the time it remains in the air, and the maximum height it attains by analyzing the projectile motion. In projectile motion, the horizontal and vertical components are treated separately.
(a) Speed at which the ball hits the ground:
Since there is no air resistance and the horizontal velocity remains constant, we can use the Pythagorean theorem to find the resultant speed when the ball hits the ground. The horizontal velocity is 14.0 m/s, and the vertical velocity upon impact will be equal in magnitude but opposite in direction to the initial vertical velocity because of symmetry in projectile motion, which is 13.0 m/s. Therefore, the speed at impact is √(14.0² + 13.0²) = 19.1 m/s.
(b) Time the ball remains in the air:
The time in the air depends only on the vertical motion. Under the influence of gravity, the ball takes the same time to go up as it does to come down. Using the equation of motion under constant acceleration (g = 9.81 m/s²), we have time t = (2 × 13.0 m/s) / g, which gives t = 2.65 s.
(c) Maximum height attained by the ball:
To find the maximum height, we use the kinematic equation for vertical motion: h = (v²) / (2 × g), where v is the initial vertical velocity. Substituting the given values, we get h = (13.0²) / (2 × 9.81) = 8.58 m.