A ball is tossed upward at 15 meters / second from a height of 12 meters above the ground.  Assume a uniform downward acceleration of 10 m/s^2 (an approximation within 2% of the 9.8 m/s^2 acceleration of gravity). 

• How high does it rise and how long does it take to get to its highest point? 

We will assume in this solution that the upward direction is positive.

This can be reasoned directly as follows:

At its maximum height the velocity of the ball is 0. 

It starts with an upward velocity of 15 m/s, and its velocity decreases by 10 m/s every second.  So it takes 1.5 seconds to reach its maximum height. 

• A little more formally, a = `dv / `dt so `dt = `dv / a = -15 m/s / (-10 m/s^2) = 1.5 s.

Its acceleration is uniform, so its average velocity on this interval is the average of its initial 15 m/s velocity and its final velocity 0 m/s. Thus its average velocity is

• vAve = (v0 + vf) / 2 = (15 m/s + 0 m/s) / 2 = 7.5 m/s. 

Rising for 1.5 s with an average velocity of 7.5 m/s is rises to height

• `ds = 7.5 m/s * 1.5 s = 11.25 meters. 

You could also use the equations of motion with v0 = 15 m/s, a = -10 m/s^2 and vf = 0. 

• The second equation of motion vf = v0 + a `dt could be solved for `dt, then the first equation `dx = (v0 + vf) / 2 * `dt would give you the displacement.

• How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground? 

Since we have already solved the question of how high the ball rises, the present question could be solved by figuring out that at the highest point it is 12 m + 11.25 m = 23.25 m above the ground, and at the initial instant of its decent it is at rest.  This would give us `ds, v0 and a for its descent. 

However this question can be solved using only the initially given information.  It is important to understand this solution.  The analysis is as follows:

We simply analyze the interval from the initial release to contact with ground.

Acceleration remains unchanged during this interval; there is no change in acceleration at the highest point. Acceleration is -10 m/s^2 throughout the ball's motion from release to first contact with the ground. 

Our initial information tells us that during this interval the ball started with upward velocity 15 m/s, accelerated at -10 m/s^2, and ended up 12 m lower than it started.  Thus

v0 = +15 m/s
`ds = -12 m
a = -10 m/s^2

Solving vf^2 = v0^2 + 2 a `ds for vf we get

vf = +- sqrt( v0^2 + 2 a `ds)
= +- sqrt( (15 m/s)^2 + 2 * (-10 m/s^2) * (-12 m) )
= +- sqrt( 225 m^2/s^2 + 240 m^2/s^2)
= +- sqrt(465 m^2/s^2)
= +-21.6 m/s, approximately.

From the specified conditions we know that the + solution is not reasonable, so we have

vf = - 21.6 m/s.

With this and the three initial quantities we can easily solve for or reason out `dt.

The simplest reasoning is that

vAve = (+15 m/s + (-21.6 m/s) ) / 2 = -3.3 m/s,

so the time interval is

`dt = `ds / vAve = -12 m / (3.3 m/s) = 3.6 sec, approximately.

Another way to solve directly for `dt is to use the same information in the third equation. This solution is a little messy and if your algebra skills arent in good shape, you might want to skip it:

`ds = v0 `dt + .5 a `dt^2 is quadratic in `dt. Putting the equation into the general form of a quadratic we get

.5 a `dt^2 + v0 `dt - `ds = 0.

This is of the form A x^2 + B x + C = 0, the standard form of a quadratic equation, where x is used in place of `dt and A = .5 a, B = v0 and C = -`ds. The solution is

x = (-B +- sqrt( B^2 - 4 A C ) ) / (2 A)
= (-v0 +- sqrt( v0^2 - 4 * (.5 a) * (-`ds) ) / (2 ( .5 a) )
= (-v0 +- sqrt( v0^2 + 2 a `ds) ) / a.

Since x was used to stand for `dt we have

`dt = (-v0 +- sqrt( v0^2 + 2 a `ds) ) / a
= (-15 m/s +- sqrt( (15 m/s)^2 + 2 * 10 m/s^2 * (-12 m) ) ) / (-10 m/s^2)
= (-15 m/s +- sqrt( 465 m^2 / s^2) ) / (-10 m/s^2)
= (-15 m/s +- 21.6 m/s) / (-10 m/s^2)
= -36.6 m/s / (-10 m/s^2) OR (+6.6 m/s) / (-10 m/s^2)
= 3.6 s OR -0.66 s.

The 3.6 s solution is consistent with the preceding solution for `dt.

The -0.66 s solution would indicate that if the projectile was already in uniformly accelerated motion prior to the initial instant, it passed through the -12 m position 0.66 s before passing through the initial point.

• At what clock time(s) will the speed of the ball be 5 meters / second? 

The speed of the ball is 5 m/s when its velocity is either 5 m/s or -5 m/s.  By direct reasoning (the velocity of the ball is decreasing by 10 m/s every second) we see that starting at 15 m/s it will attain velocity 5 m/s after 1 second, and -5 m/s after 2 seconds.  The same results could be obtained solving the equation vf = v0 + a `dt for `dt. 

• Using vf = 5 m/s, v0 = 15 m/s, a = -10 m/s we find that `dt = (vf - v0) / a is 1 second. 

• Using vf = -5 m/s, v0 = 15 m/s, a = -10 m/s we find that `dt = (vf - v0) / a is 2 seconds. 

• At what clock time(s) will the ball be 20 meters above the ground?

We first must specify where the ball is at clock time t = 0.  We will make the common assumption that the t = 0 instant coincides with the release of the ball.

To get to the 20 meter height its vertical position much change by 8 meters (remember that the ball started at the 12 meter height). 

So we can solve for the time interval between release and the 20 meter height, using

• `ds = 8 m,

• a = -10 m/s^2 and

• v0 = 15 m/s. 

This situation resists direct reasoning, so we use the fourth equation of motion:

vf^2 = v0^2 + 2 a `ds, so

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

+-sqrt( (15 m/s)^2 + 2 * (-10 m/s^2) * 8 m) =

+- sqrt( 65 m^2 / s^2) = +-8.1 m/s, approx.. 

If the velocity is 8.1 m/s, the change in velocity is thus -6.9 m/s, which requires .69 s. 

• (This could also be obtained from the original information using the third equation of motion; however that equation is quadratic in `dt.  University physics students should have no trouble with the quadratic; principles of physics and general college physics students whose algebra background is sufficient are also welcome to use this approach). 

• (Another approach would be to use v0, vf and a in the second equation of motion, solving for `dt; we would obtain `dt = (vf - v0) / a = (8.1 m/s - 15 m/s) / (-10 m/s^2) = .69 s).

If the velocity is -8.1 m/s, which occurs on the way down, then solving the second equation of motion for `dt we obtain

`dt = (vf - v0) / a = (-8.1 m/s - 15 m/s) / (-10 m/s^2) = 2.31 s.
 

• How high will it be at the end of the sixth second?

Of course the ground will probably intervene, causing acceleration to become nonuniform and rendering this analysis meaningless.

On the other hand there could be a deep hole in the ground (e.g., a well shaft).

In this event you could reason as follows:

After 6 seconds the ball has velocity 15 cm/s + (-10 m/s^2) * 6 s = -45 m/s.

Its average velocity for the 6 seconds is thus (15 m/s + (-45 m/s) ) / 2 = -15 m/s, and its height relative to its release position is -15 m/s * 6 s = - 90 m.

This puts it 78 m below the level of the ground.