Final answer:
The ball reaches a maximum height of 20 m and takes 2 seconds to do so. It takes a total of 4 seconds to hit the ground again, and halfway to the maximum height, the velocity is 14.14 m/s.
Step-by-step explanation:
Calculations for a Ball Thrown Vertically Upward
To calculate the maximum height reached by a ball thrown vertically upward with an initial velocity of 20 m/s, we use the kinematic equation (v^2 = u^2 + 2as), where v is the final velocity (0 m/s at the maximum height), u is the initial velocity (20 m/s), a is the acceleration due to gravity (-10 m/s^2 because it is in the opposite direction to the motion), and s is the distance (maximum height). Solving for s gives us s = (v^2 - u^2) / (2a), which results in a maximum height of 20 m.
To find out the time taken to reach the maximum height, we use (v = u + at). Rearranging for t gives us t = (v - u) / a, resulting in a time of 2 seconds to reach the maximum height.
The total time taken to reach the ground again will be twice the time taken to reach the maximum height, because the upward and downward journeys are symmetrical. Consequently, the ball will take a total of 4 seconds to reach the ground.
Lastly, to calculate the velocity reached halfway to the maximum height, we first find the halfway height, which is 10 m. Using the kinematic equation (v^2 = u^2 + 2as) where s is now 10 m, and solving for v, we get a velocity of 14.14 m/s.