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).
• 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: 15 m/s / 10 m/s^2 = 1.5s
so it takes 1.5 s to reach its highest point.
We then consider the time interval in which it took the ball to reach its maximum height. To find the max height, we calculate the position of the ball when its velocity equals zero (i.e. max height). At t=0 we have y0 = 0, v0 = 15 m/s, and a = -10 m/s^2. At time t (max height), v = 0, a = -10 m/s careful; units are m/s^2
so we want to find y.
V2 = vo2 + 2ay. Then solve for y.
Y = (v2 – vo2) / 2a = 0 – (15 m/s)^2 / 2 (- 10 m/s^2) = 11.25 m, this is the highest point
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15 min
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You've answered the first question, and you did very well on that question.
Can you answer the rest?