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A ball rolls off the end of an incline with a vertical velocity of 20 cm/s downward, and a horizontal velocity of 80 cm/s. The ball falls freely to the floor 120 cm below.

For the interval between the end of the ramp and the floor, what are the ball's initial velocity, displacement and acceleration in the vertical direction?

What therefore are its final velocity, displacement, change in velocity and average velocity in the vertical direction?

Short synopsis of reasoning:

20 cm/s downward is the initial vertical velocity. The vertical velocity changes due to the acceleration of gravity.

The horizontal velocity is not affected by the downward force of gravity, and it remains constant at 80 cm/s.

So in the vertical direction, taking downward as the positive direction, v0 = 20 cm/s, a = 980 cm/s^2 and `ds = 120 cm. From this you can find `dt.

In the horizontal direction, velocity is constant so you can use the definition of average velocity along with the `dt you found for the vertical motion, in order to find `ds in that direction.

Detailed solution:

The vertical motion is independent of the horizontal. 

We first solve for the vertical motion:

The vertical motion is characterized by a downward acceleration of 980 cm/s^2, a downward displacement of 120 cm and an initial downward velocity of 20 cm/s.  Thus for the vertical motion, if we choose the downward direction to be positive, we have

v0 = 20 cm/s, a = 980 cm/s^2 and `ds = 120 cm.

We cannot reason anything directly from this information. 

Both the third and fourth equations of uniformly accelerated motion include the three variables v0, a and `ds.  So the third equation could be solved for `dt, or the fourth for vf.

It is not recommended that students without a strong mathematical background solve the third equation for `dt; the equation is quadratic in `dt and the solution is algebraically challenging.  University Physics students must have a strong mathematical background and should be able to solve the quadratic and interpret the solutions.

So we use the fourth equation vf^2 = v0^2 + 2 a `ds.  We solve this equation for v0, obtaining

v0 = + - sqrt(v0^2 + 2 a `ds).

Substituting our values for v0, a and `ds we have

vf = +- sqrt( (20 cm/s)^2 + 2 * 980 cm/s^2 * 120 cm) = +- 480 cm/s.

The final velocity is clearly in the downward direction, so we discard the solution -480 cm/s and accept the solution + 480 cm/s.

Initial velocity v0 and final velocity vf, with uniform acceleration, imply average velocity

vAve = (v0 + vf) / 2 = (20 cm/s + 480 cm/s) / 2 = 250 cm/s

so that the time required for the fall is

`dt = `ds / vAve = 120 cm / (250 cm/s) = .48 sec.

What are the ball's acceleration and initial velocity in the horizontal direction, and what is the change in clock time, during this interval?

The horizontal motion of an ideal projectile is characterized by zero acceleration (since the only force acting on an ideal projectile there the vertical gravitational force, it experiences no force in the horizontal direction, implying zero acceleration in that direction). 

The initial horizontal velocity is 80 cm/s, and since acceleration is zero this velocity remains unchanged.  Thus the average horizontal velocity is also 80 cm/s and we conclude that the horizontal displacement is therefore

After the instant of impact with the floor, can we expect that the ball will be uniformly accelerated?

Upon impact with the floor the ball and the floor both begin compressing, exerting equal and opposite forces on one another. 

With increasing compression the force between the floor and the ball will increase. 

Why does this analysis stop at the instant of impact with the floor?

This analysis assumes uniform acceleration, and acceleration is no longer uniform. 

INCORRECT STUDENT SOLUTION AND INSTRUCTOR COMMENTARY

v0= 20cm/s , vf = 0 , `ds = 120cm,

Thus `dv = -20cm/s, `dt = `ds/vAve = 120cm / -10cm/s = 12s

You assumed final vertical velocity zero; with this assumption your analysis would be correct, but this assumption is not compatible with uniformity of acceleration.

During its uniform acceleration to the floor the ball attains a downward vertical velocity of nearly 500 cm/s. Upon contact with the floor acceleration becomes nonuniform and is not included in the analysis. 

For the vertical motion:

Using the fourth equation of motion vf^2 = v0^2 + 2 a `ds, along with the vertical quantities v0 = 20 cm/s, `ds = 120 cm and a = 980 cm/s^2 (all quantities are in the downward direction so by these choices we have implicitly designated downward as positive), we get

vf = +-sqrt(v0^2 + 2 a `ds) = +- sqrt( (20 cm/s)^2 + 2 * 980 cm/s^2 * 120 cm) = +- sqrt(25 000 cm^2 / s^2) = +-500 cm/s (all calculations approximate).

We reject the negative result.

What are the ball's acceleration and initial velocity in the horizontal direction, and what is the change in clock time, during this interval?

v0 = 80cm/s , vAve = 80cm/s / 2 = 40cm/s

Thus `dt = 120cm*40cm/s = 4.8s

The final velocity in the horizontal direction, for the uniform-acceleration interval being analyzed, is not zero.  The velocity is constant, so 80 cm/s is the average velocity (as well as the initial and final velocity). 

So for the horizontal direction: