phy 201
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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?
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 what clock time(s) will the speed of the ball be 5 meters / second?
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: ->->->->->->->->->->->-> :
acceleration = final velocity - initial velocity / time
time = 15m/10m/s= 1.5sec
maximium height = velocity * time
= 15m/s *1.5 = 22.5m
time at 5m/s = 22.5 / 5m/s = 4.5sec
time at 20m above ground = 2om/ 15m/s = 1.33sec
at the end of six second
distance = 15m/s*6sec = 90 m
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approx 15 minutes
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Good, but be sure you understand how to use positive and negative quantities to represent everything. No need to resubmit if you understand:
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Solution
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