Phy 231
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?
15 m/s - 10 m/s^2 *'dt = 0 m/s, which is the highest point; therefore, (-15 m/s)/(-10 m/s^2) = 'dt ; 1.5 s = dt. Now we plug this back into the distance equation to find how far it rises 12 m + 15 m/s * 1.5s - (.5)10 m/s^2 * 1.5^2 = 'ds ; 12 m + 22.5 m - (.5)* 22.5 = 'ds ; 12 m + 11.25 m = 'ds ; 23.25 m = 'ds in 1.5 s
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 final position of 0 we find that 0 m = 23.25 m – 5 m/s^2 'dt^2; therefore, -23.25 m = - 5 m/s^2 *'dt^2 ; 4.65 s^2 = 'dt^2 and by taking the square root we find that 2.15 s = 'dt is how long it takes for it to reach to ground after reaching maximum height. We now use this time to find final velocity at impact as 10 m/s^2 * 2.15 s = 21.5 m/s, and the total time necessary from the initial toss would be 1.5 s + 2.15 s = 3.65 s.
At what clock time(s) will the speed of the ball be 5 meters / second?
At clock times of 1 s when 15 m/s – 10 m/s^2 ( 1s) = 5 m/s, and at clock time when 5 m/s = 10 m/s^2 'dt ; .5 s = 'dt after maximum height; therefore the second clock time would be .5 + 1.5 s = 2 s after the ball was tossed.
At what clock time(s) will the ball be 20 meters above the ground?
Distance of 20 m = 12 m + 15 m/s 'dt – 5 m/s^2 'dt^2 ; 0 = -5 m/s^2 * 'dt^2 + 15 m/s – 8 m now by using the quadratic formula we can find the two clock times that give us the distance of 20 m. [ - 15 +- sqrt(225 – (4)(-5)(-8))]/(2)(-5) = [ -15 +- sqrt( 225 – 160)]/-10 = [ -15 +- sqrt(65)]/ -10 = (-15 +- 8.06)/-10 = -23.06/ -10 & -6.92/ -10 = 2.306 & .692 approximately.
How high will it be at the end of the sixth second?
answer/question/discussion: Since it would only take 3.65 seconds after being tossed to reach the ground, at the clock time of 6 seconds it would still be at the ground position of 0 m.
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25 minutes
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You did have the ground on this one. Good solution.