Phy 201
Your 'cq_1_8.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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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).
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How high does it rise and how long does it take to get to its highest point? y = (v^2 - vo^2) / 2(-10 m/s^2) = -225 m/s / -20 m/s^2 = 11.25 m (from the hand) + 12 m = 23.25 m from the ground; 0 = 15 m/s * t + 1/2(-10 m/s^2) * t^2 = t=0 & t = 3 s; half of the total time will give 1.5 seconds to reach maximum height
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The units would be -225 m^2/s^2 / (-20 m/s^2) = 11.25 m. Your units, (m/s) / (m/s^2), would yield s, not m.
How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground? v = 15 m/s + (-10 m/s^2)(4.25 s) = -27.5 m/s; 15 m/s / 12 m = 1.25 s + 3 s = 4.25 s
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See the given solution below.
At what clock time(s) will the speed of the ball be 5 meters / second? t = (v - vo) / a = (5 m/s - 15 m/s) / -10 m/s^2 = 1 s; given that y=0 at 1.5 s, then at 2 s the velocity will be 5 m/s as well.
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At what clock time(s) will the ball be 20 meters above the ground? using the quadratic formula we get : 15 m/s +/- sqrt[(15 m/s)^2 - 4(-5 m/s^2)(20 m)] / 2(-5 m/s^2) = t = -3.5 s
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How high will it be at the end of the sixth second?
answer/question/discussion: buried
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35 minutes
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Your solution doesn't appear to be totally correct. Check against the following and let me know if you have questions:
It isn't necessary to solve for or use the maximum height (which in this problem was previously requested) in order to find the result requested here. You can start with the given initial information and measure `ds with respect to the initial point. The analysis is as follows:
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. 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 indicates 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 instant.