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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 approxomation 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?
• How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground?
• At what clock time(s) will the speed of the ball be 5 meters / second?
• At what clock time(s) will the ball be 20 meters above the ground?
• How high will it be at the end of the sixth second?
answer/question/discussion:
v0=15m/s a=-10m/s^2
‘ds = (v^2-v0^2)/2a=0(-(15m/s)^2)/2*-10m/s^2=-225m^2/s^2/-20m/s^2=11.5m is the rise and T=-v0/a=-15m/s/-10m/s=1.5s is the time it takes to reach its highest point
12m + 11.5m = 23.5m is the total distance the ball has to travel to the ground. Vo would be 0m/s since the measurement is starting from the beginning of the fall in which velocity is 0m/s. Vf^2 = v0^2 +2a’ds; vf^2 = 2(10m/s^2)23.5m = sqrt(470m^2/s^2)=22m/s. vf=v0+a*’dt =(vf-v0)/a = ‘dt =( 22m/s – 0m/s)/10m/s^2 =2.2s = ‘dt fall ‘dt total = rise time +fall time = 1.5s+2.2S= 3.7s is the time it takes the ball to strike the ground after the initial fall.
Vf=5m/s and -5m/s, v0 = 15m/s, a = -10m/s^2 ‘dt = (vf-v0)/a =1s when ball is rising and 2s when all is falling.
Since the starting point is 12m then the ‘ds to 20m is 8m. using the formula ‘ds = v0*’dt +.5a’dt^2; 8 =(15m/s)t +.5(-10m/s^2)t^2; to solve in the quadratic formula use the for at^2 +bt +c =0 where a is not acceleration. (5m/s^2)t^2 + (15m/s)t +8m = 0 the quadratic formula will be t = 15m/s +- sqrt((15m/s)^2 -4(5m/s^2)*8m)/2(5m/s^2 = .7s and 2.2s
The ball should be on the ground at rest by the time duration of six seconds
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:
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.
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.75 hours
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Would like to have feedback. Thanks, Joshua
&#Good work. See my notes and let me know if you have questions. &#