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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Assn 8 Seed 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 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?
`ds= (vf^2-v0^2)/2a, so `ds = (0^2-15^2)/2*(-10)= (0-225)/-20= -225/-20= 11.25m from start point, 23.25m off ground
`dt= (vf-v0)/a= -15/-10= 1.5s
How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground?
vf^2= v0^2 + 2*a*`ds, so vf^2= 0^2 + 2(-10)(-23.25)= 0+465 => vf^2= 465, so `sqrtvf^2 = `sqrt465=> vf = 21.56m/s
`dt= (vf-v0)/a, so 21.56/10= 2.156s
Total Time = 1.5s+2.156= 3.656s
At what clock time(s) will the speed of the ball be 5 meters / second?
(vf-v0)/a= `dt, so (5-15)/-10 = 1s
5m/s @ 1 sec and 2sec
At what clock time(s) will the ball be 20 meters above the ground?
t= 15+-`sqrt(15^2-4*5*8)/2(5)= 15+-`sqrt(225-160)= 15+-`sqrt65
2.3s & .6s
How high will it be at the end of the sixth second?
answer/question/discussion:
`ds= v0 * `dt + .5a*`dt^2, so `ds= 15*6 + (-5)36= 90-180= -90m
I guess this is figuring the object penetrates the ground?
With that implicit assumption this is correct. The assumption is appropriate.
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45min
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The clock time questions threw me off course a bit and I had to brush up on the quadratic equation.
You handles everything very well. Excellent work.