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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?
The initial velocity is 15m/sec and decreases by 10m/sec until t = 2.5 seconds when the velocity is 0. I calculate the average velocity to get my distance.
It only takes 1.5 seconds for a 15 m/s initial velocity to reach 0 m/s when accelerating at -10 m/s^2.
vAve = (v0+vf)/2
vAve= (15 + 0)/2
vAve= 7.5 m /sec
‘ds = vAve * ‘dt
‘ds =7.5m/sec * 2.5sec
‘ds = 18.75
However, the ball started at 12 meters above the ground so I need to add 12 to my answer.
The ball is 30.75 meters above the ground.
• How fast is it then going when it hits the ground, and how long after the initial
• toss does it first strike the ground?
To find the final velocity, I start where the ball’s initial velocity is 0m/sec and it has to travel 30.75 meters to hit the ground. I use one of the four equations to find final velocity.
vf^2= v0^2 + 2a*’ds
vf^2 = 2a*’ds
vf^2 = 2(10m/sec^2) * 30.75m
vf^2 = 20m/sec^2 * 30.75m
vf^2 = 615m^2/sec^2
vf = 24.8 m/sec
Your solution has the ball going higher than it actually does, but otherwise this would work. However to answer this question it is unnecessary to use your solution for the maximum height:
Just 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 before and after the highest point.
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.
`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.
• At what clock time(s) will the speed of the ball be 5 meters / second?
At t=1 the ball decreases to 5 m/sec. It is also traveling at 5m/sec at t=3, however in the negative direction.
• At what clock time(s) will the ball be 20 meters above the ground?
Assuming all of my other answers are correct, the ball will be 20m above the ground at about t =3.5 seconds. At t = 2.5 seconds, the ball is 30.75 meters above the ground, so after another second the ball will be about 20.75 meters above the ground, which is close enough to 20 meters,
You would have two solutions, as in previous problems. In this case both solutions have meaning.
The mathematics matches the reality; we find that, as in common experience, the ball passes the 20 meter position going up and coming down.
For the interval between release and 20 m height we can use the following:
The 20 m height occurs when the ball is 8 m above the release point. (Note that the ground is the only point from which the ball will rise as high as 20 m; this will occur when the ball is 8 m above the release point).
v0 = 15 m/s
`ds = 8 m
a = -10 m/s^2
We can use either `ds = v0 `dt + .5 a `dt^2 or vf^2 = v0^2 + 2 a `ds to solve this problem.
Using the latter we would obtain
vf = +- sqrt(v0^2 + 2 a `ds) = +- sqrt( (15 m/s)^2 + 2 (-10 m/s^2)(8 m) ) = +- sqrt(225 m^2 / s^2 - 160 m^2 / s^2) = +- sqrt(65 m^2/s^2) = +- 8.1 m/s, approx..
If vf = +8.1 m/s, then `ds = (vf - v0) / a = (8.1 m/s - 15 m/s) / (-10 m/s^2) = .69 s.
If vf = -8.1 m/s, then `ds = (vf - v0) / a = (-8.1 m/s - 15 m/s) / (-10 m/s^2) = 2.31 s.
The ball passes the 20 m height at t = .69 s moving at +8.1 m/s (it's therefore on the way up), and at t = 2.31 s moving at -8.31 m/s on the way back down. Note that these instants are symmetric with respect to the t = 1.5 s instant at which the ball reaches its maximum vertical position.
• How high will it be at the end of the sixth second?
The ball is expected to hit the ground at about t = 5.5 seconds. At the sixth second the ball will remain on the ground.
answer/question/discussion:
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 wellshaft).
In this case you would find that the ball has velocity 15 cm/s + (-10 m/s^2) * 6 s = -45 m/s. Its average velocity for the 3 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.
If there's a hole in the ground (maybe a wellshaft) this is possible. Ordinarily, though, we would expect the balls' contact with the ground to end the uniform-acceleration phase prior to its reaching this position, and our analysis beyond that instant would be irrelevant to the ball's actual motion.
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25 minutes
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Be sure to see my notes and let me know if you have questions.