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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?
The ball rises 11.25m and it takes 1.5s to get to its highest point.
`dt = (vf - v0)/a = (0 - 15)/-10 = 1.5s
`ds = vf^2 - v0^2/2a = (0) - (15)/2(-10) = -225/-20 = 11.25m
How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground?
The ball is traveling -21.6m/s when it hits the ground and 3.66s have elapsed since the initial toss and when the ball first strikes the ground.
Max. height = 12m + 11.25m = 23.25m
vf^2 = v0^2 + 2a`ds = (0m/s) + 2(-10)(-23.25) = 465m^2/s^2
`sqrt vf^2 = `sqrt 465m^2/s^2
vf = +- `sqrt 465 = +-21.5683m/s = -21.56 = -21.6m/s
`dt = (vf - v0)/a = (21.56 - 0)/-10 = -2.156 = 2.156s
2.156s + 1.5s = 3.66s
At what clock time(s) will the speed of the ball be 5 meters / second?
1s and 2s
At what clock time(s) will the ball be 20 meters above the ground?
1.07s and 2.31s
How high will it be at the end of the sixth second?
0m (on the ground)
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35min.
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Please respond relative to goal of C.
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