PHY201
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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Seed Question 8.2
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?
Y = (v^2 – vo^2)/ 2a = (0 – 15m/s^2) / 2*-10m/s/s = -15m/s^2 / -20m/s/s = .75m
T= -v0/a = -15m/s / -10m/s/s = 1.5s
-15 m/s^2 is not a quantity associated with this problem. (0^2 - (15 m/s)^2) = -225 m^2/s^2 would be associated with the problem, and would lead you to a correct answer.
You should probably avoid saying just that t = -v0 / a. This is not generally true, though in this case with v0 = 0 it is. You should either preface the equation with a brief statement such as 'since vf = 0 ... ' or equivalently say that t = `dv / a = (vf - v0) / a = (0 - v0) / a = -v0 / a.
• 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 = v0 + at= 15m/s + (10m/s/s * 1.5s) = 15 +15 = 30m/s
v0 and a have opposite signs, one being directed upward and the other downward.
You cannot assume equal times up and down, and even if you do this result would be incorrect.
The ball is released from a .12 m above the ground, motion up is not symmetric with motion down. As it passes the 12 m point on its way down, its velocity is in fact -15 m/s and the time interval would be calculated as `dt = `dv / a = (vf - v0) / a = (-15 m/s - 15 m/s) / (-10 m/s^2) = 3 s.
However it's not yet at the ground when `dt = 3 s.
T = v/a = 30m/s / 10m/s/s = 3s
• At what clock time(s) will the speed of the ball be 5 meters / second?
T = (v-v0)/a = (5m/s – 15m/s) /-10m/s/s = -10/-10 = 1s
T = (-5m/s – 15m/s) / -10m/s = -20/-10 = 2s
• At what clock time(s) will the ball be 20 meters above the ground?
Y = y0 + v0t + 1/2at^2
20m = 12m + (15m/s)t + ½(-10m/s/s)t^2
-5m/s/st^2 – 15m/st + 8m = 0
T =( 15m/s +/- sqrt((15m/s)^2 – (4*-5m/s/s * 8m))/(2*-5m/s/s)
T= .46 and T = .14
Good, but the final results don't make sense. Even if it didn't slow down, and upward velocity of 15 m/s wouldn't get you there in .14 s. Your equation actually gives you two results, one positive than one negative. This is certainly possible with a quadratic equation and occurs frequently, but in this case it is clear that the ball passes the 20 m mark going up then coming down, both after the initial instant, so the solutions both need to be positive.
Your equation should be
-5m/s/st^2 + 15m/st - 8m = 0.
If you then do the arithmetic correctly you should get good results.
• How high will it be at the end of the sixth second?
answer/question/discussion:
v = v0 + at = 15m/s + -10m/s/s * 6s = 15m/s – 60m/s = -45m/s
y = v0t + ½ at^2 = 15m/s * 6s + ½ *-10m/s/s * 6s^2 = 90 + ½ * -360 = 90- 180 = -90m
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1 hour
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&#Good responses. See my notes and let me know if you have questions. &#