An elevator without a ceiling is ascending with a constant speed of 10m/s. A boy on the elevator shoots a ball directly upward, from a height of 2.0m above the elevator floor, just as the elevator floor is 28m above the ground. The initial speed of the ball with respect to the elevator is 20m/s.

a) How long does the ball take to return to the elevator floor?

To find out how long the ball takes to return to the elevator floor, we need to calculate the time it takes for the ball to reach its highest point and then come back down to the floor.

Let's break down the problem step by step:

1. First, let's find the time it takes for the ball to reach its highest point. We can use the formula for the time of flight of an object thrown vertically:

time of flight = 2 * (initial vertical velocity) / (acceleration due to gravity)

In this case, the initial vertical velocity of the ball is 20 m/s (upward) and the acceleration due to gravity is 9.8 m/s² (downward). Therefore:

time of flight = 2 * 20 m/s / 9.8 m/s²

2. Next, we need to calculate the total distance that the ball travels from the elevator floor to its highest point. The ball starts 2.0 m above the elevator floor and has to travel an additional distance equal to the height of the elevator floor above the ground, which is 28 m. So the total distance is:

total distance = 2.0 m + 28 m

3. Now we can find the time it takes for the ball to reach the highest point using the formula for time:

time = total distance / (initial vertical velocity)

Therefore:

time = (2.0 m + 28 m) / 20 m/s

4. Finally, to find the total time for the ball to return to the elevator floor, we need to double the time it took to reach the highest point (since the time to come down is the same as the time to go up).

total time = 2 * time

Now you can plug in the values and calculate the time it takes for the ball to return to the elevator floor.

0.2227

The elevator is not accelerating. Therefore you and the ball can not feel how fast it is going. The problem is just the same as if you did the experiment on the ground. The 10 m/s is irrelevant and the 28 m above ground is irrelevant and the problem is

h = 0 = 2 + 20 t - (1/2)(9.8) t^2
solve the quadratic for t

the answer lies in the book....

peace and wisdom