Final answer:
At the top of its path, a ball tossed upwards at 28 m/s has a velocity of 0 m/s and an acceleration of 9.8 m/s^2 downward. It takes approximately 2.86 seconds to reach this point, and the top is around 40 meters from the hand. The velocity just before it is caught is -28 m/s, the total displacement is 0 m, and the total time for the motion is about 5.72 seconds.
Step-by-step explanation:
When a ball is tossed upwards at 28 m/s, it will slow down until it reaches the top of its path. At the top, the velocity of the ball is 0 m/s because it momentarily stops before starting to fall back down. The acceleration due to gravity is always acting downwards at approximately 9.8 m/s2, regardless of the ball's position in its path.
To find the time it takes for the ball to reach the top, we can use the equation v = u + at, where v is the final velocity (0 m/s at the top), u is the initial velocity (28 m/s upward), a is the acceleration due to gravity (-9.8 m/s2, negative because the acceleration is downward), and t is the time. Solving for t gives us t = (v - u) / a, so t = (0 - 28) / -9.8, which is approximately 2.86 seconds.
The height reached by the ball can be calculated using the equation s = ut + 0.5at2, where s is the displacement. Substituting the known values gives us s = 28 * 2.86 + 0.5 * (-9.8) * (2.86)2, resulting in a height of approximately 40 meters.
The total displacement of the ball is 0 m since it is caught at the same height from which it was tossed. The velocity just before being caught will be equal in magnitude but opposite in direction to the initial velocity, so -28 m/s. The total time for the motion will be double the time taken to reach the top, which is roughly 5.72 seconds.