Final answer:
The ball hits the ground at a speed of 22.8 m/s. It remains in the air for 3.27 seconds. The maximum height attained by the ball is 52.5 meters.
Step-by-step explanation:
(a) To find the speed at which the ball hits the ground, we can use the fact that the horizontal and vertical motions of the ball are independent. The initial horizontal velocity remains constant throughout the motion, so it does not affect the speed at which the ball hits the ground. The vertical component of the motion is influenced by gravity, so we need to find the time it takes for the ball to reach the ground. We can use the vertical motion equation y = v0yt - (1/2)gt2 where y is the vertical displacement, v0y is the initial vertical velocity, t is the time, and g is the acceleration due to gravity. Since the ball is kicked from the ground, y will be zero and we can solve for t. Rearranging the equation to solve for t, we have t = sqrt(2y/g). Plugging in the values, t = sqrt(2*0/9.8) = 0 s. Therefore, the ball hits the ground instantaneously in the horizontal direction and with a vertical velocity of 16 m/s. The horizontal and vertical components of the velocity can be combined using the Pythagorean theorem to find the speed at which the ball hits the ground, which is sqrt((17.0 m/s)^2 + (16.0 m/s)^2) = 22.8 m/s.
(b) The time the ball remains in the air can be found using the vertical motion equation y = v0yt - (1/2)gt2, where y is the vertical displacement, v0y is the initial vertical velocity, t is the time, and g is the acceleration due to gravity. The initial vertical velocity is 16 m/s and the ball reaches the ground when y is zero. Plugging in these values, we have 0 = (16 m/s)t - (1/2)(9.8 m/s^2)t^2. Rearranging the equation and solving for t, we have t = (2v0y)/g = (2*16 m/s)/(9.8 m/s^2) = 3.27 s. Therefore, the ball remains in the air for 3.27 seconds.
(c) The maximum height attained by the ball can be found using the vertical motion equation y = v0yt - (1/2)gt^2. We know that the initial vertical velocity (v0y) is 16 m/s and the time (t) can be found using the equation t = (2v0y)/g = (2*16 m/s)/(9.8 m/s^2) = 3.27 s. Plugging these values into the equation for y, we have y = (16 m/s)(3.27 s) - (1/2)(9.8 m/s^2)(3.27 s)^2 = 52.5 m. Therefore, the maximum height attained by the ball is 52.5 meters.