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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With v0 = 15m/s^2, a = -10m/s^2, we use t = (v-v0)/a, therefore, t = (0-15m/s)/-10m/s^2 => t = -15m/s / -10m/s^2 => t =1.5s. The ball reaches a max height after 1.5 seconds.
Using y = (v^2 – v0^2)/2a => y = -225m^2/s^2 / -20m/s^2 => y = 11.25m. The ball reaches the max height of 11.25 meters after 1.5 seconds.
Starting at the max height, v0 = 0 and (x-x0) = 23.25m, using v^2 = v0^2 + 2a(x-x0) => v^2 = 465m^2/s^2 => v = 21.6m/s. The ball has a velocity of 21.6m/s when the ball hits the ground.
Using t = (v-v0)/a, t = 21.6m/s / 10m/s^2 => t = 2.16s, the ball hits the ground 2.16s after it reached its max height, therefore, the ball hits the ground after a total of 2.16s + 1.5s = 3.66s.
The ball will reach a speed of 5m/s at clock time of 1s, using t = (v – v0)/a, substituting v = 5m/s, v0 = 15m/s, and a = -10m/s^2. It will also reach speed of 5m/s at clock time of 2s, .5 secs after its max height. Using v = v0 + at, solving for t, t = (v – v0)/a, substituting v = 5m/s, v0 = 0 (at max height), and a = 10m/s^2. We get t = .5s.
Using the quadratic formula and the equation y = y0 + v0*t + .5*a*t^2 with t = 0, v0 = 15m/s, and y = 20m.
20m = 12m + (15m/s)t + (-5m/s^2)t^2, rewriting we get (5m/s)t^2 + (-15m/s)t + 8m. Using the quadratic formula and substituting we get t = (15m/s +-sqrt(65m^2))/10m/s^2 => t = .7s and 2.3s
The ball reaches a height of 20m at t = .7s and t = 2.3s.
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30mins
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Everything you're doing here looks good, but I can't tell for sure if you're answering all questions correctly unless you insert your answers into a copy of the questions. I suspect everything is correct; it's your option if you want to resubmit for my further scrutiny.