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phy 231
Your 'cq_1_08.2' 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?
answer/question/discussion: ->->->->->->->->->->->-> :
`dv must be -15 m/s in order for the ball to be at its highest point vf = 0.
We also know vAve = (15 m/s + 0)/2 = 7.5 m/s and that a = -10 m/s/s
Using vf = v0 + a*`dt and solving for `dt:
`dt = (vf - v0)/a = (0 - 15 m/s)/-10 m/s/s = 1.5 seconds
To find `ds, use `ds = vAve * `dt = 7.5 m/s * 1.5 sec = 11.25 meters
The total height would be `ds + 12 meters = 23.25 meters
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• How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground?
answer/question/discussion: ->->->->->->->->->->->-> :
If it takes 1.5 sec to reach its peak, then it will take 1.5 sec to return to where it started. The ball will still travel 12 meter further to the ground. At this point 3 seconds has elapsed. The velocity at this point is found by vf = v0 + a*`dt = 15 m/s - 10m/s/s*3s = 15 - 30 = -15 m/s. Now we have a ball with velocity -15 m/s falling to the ground in 12 meters. `dt, the time to fall, is 12m/15 m/s = .8 sec
The total time elapsed is now 3 + .8 = 3.8 seconds.
@& The ball doesn't fall the last 12 meters at a constant, or an average, velocity of 15 m/s. The ball is speeding up, and takes less than .8 seconds to cover this distance.
From the initial instant to the instant of contact with the ground the displacement is -12 m, the acceleration 9.8 m/s^2 and the initial velocity 15 m/s. That, with the equations of motion, is sufficient to obtain two possible values for the final velocity, and two possible values for the time interval. Only one of the velocities and one of the time intervals will make sense.*@
Now we can find the vf when it hits the ground. vf = v0 + a*`dt = -15 m/s - 10m/s^2*0.8sec = -15-8 = -23 m/sec
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• At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
Through reasoning, if the ball leaves at 15 m/s and is being pulled by a 10 m/s^2 acceleration of gravity, after 1 sec the ball has a 15 - 10 = 5 m/s velocity.
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• 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: ->->->->->->->->->->->-> :
The ball hits the ground in 3.8 seconds, so the sixth second values are not relevant.
20 meters = 15 m/s^2 * `dt + ½ (-10 m/s/s)*(`dt)^2 + 12, this is a quadratic equation that can be rewritten as y = -5t^2 + 15t - 8. Using the quadratic formula yields two times, 2.3 seconds and .694 seconds, which tells us the ball is at 20 meters above ground going up at .694 sec and coming down at 2.3 seconds.
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@& Good work, except for one incorrect assumption. See if you can, within a reasonable time frame (15-20 minutes), figure that one out. If not, submit your best thinking and I'll give you a link to a broader discussion.*@