Final answer:
To find the number of balls thrown up per second where each reaches a height of 5 meters, we use kinematic equations to determine the initial velocity and time to reach the maximum height. The calculations show that one ball is thrown per second.
Step-by-step explanation:
Finding the Number of Balls Thrown Up Per Second
To determine the number of balls thrown up per second, with each ball being thrown when the previous one reaches a maximum height of 5 meters, we can use the kinematic equations of motion under the influence of gravity. The acceleration due to gravity is -9.81 m/s2 and is constant near the Earth's surface. Since the balls are thrown vertically and reach a maximum height, their final velocity at that height is 0 m/s.
Using the kinematic equation v2 = u2 + 2as, where 'v' is the final velocity, 'u' is the initial velocity, 'a' is the acceleration, and 's' is the displacement, we have:
- 0 = u2 - 2(9.81)(5)
- u2 = 2(9.81)(5)
- u = √(2 * 9.81 * 5)
- u ≈ 9.9 m/s
Knowing the initial velocity, we can now find the time taken to reach the maximum height using v = u + at, solving for 't' we get:
- 0 = 9.9 - 9.81t
- t ≈ 1 second (to reach maximum height)
Since each ball is thrown at the moment the previous one reaches its maximum height, which takes about 1 second, we can infer that the number of balls thrown per second is 1 ball per second.