A ball is thrown directly downward at 4 m/s from the edge of a high cliff. If needed, use g = 10 m/s2. With zero air resistance, the distance the ball falls during its third second of fall is most nearly 28 m
4 m
29 m
12 m
57 m

57 meters

Explanation:

Let's see what variables we have in this problem:

Let's set the upwards direction to be positive and the downwards direction to be negative.

We are given the initial velocity, 4 m/s. Since the ball is thrown directly downward, we can say that the initial velocity = -4 m/s.

We are also given the acceleration due to gravity, and since the acceleration is pointing downwards, we can say that a = -10 m/s².

The time is also given to us; the question wants to know the vertical displacement at time = 3 seconds, so we can plug in 3 seconds for t.

Since we are solving for vertical displacement, we can use as one of our variables.

Now we have all of the variables except for the final velocity, .

Since we are dealing with constant acceleration, we can use this constant acceleration equation:

Subtract from both sides to get .

Substitute in the known values:

Get rid of the units to make the equation more readable.

Simplify the equation and solve for delta x.

The vertical displacement is -57 meters, so we can say that the distance the ball falls in 3 seconds is most nearly 57 meters.

im guessing the west of station b