cq_1_082

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?

answer/question/discussion: ->->->->->->->->->->->-> : We have a decrease acceleration of 10m/s and an initial velocity of 15m/s. `dt = (vf – v0)/a `dt = (0m/s – 15m/s)/-10m/s^2 = 1.5 seconds.

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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 at 1.5 seconds we are at a velocity of zero, then our ball will start coming back down at this point. At the 1.5 second mark we have increased our distance from the ground by 22.5m/s

22.5 m/s isn't a height; that's probably just a typo but be careful with your units

22.5 m is the height you would attain moving at 15 m/s for 1.5 s; however while the ball starts out at 15 m/s it immediately beings slowing

giving us a total height of 34.5 meters. Now we need to calculate how long it will take to reach 0 meters. If our displacement is 34.5 meters then we can divide that by our acceleration to calculate the seconds needed to reach the ground. This gives us 3.45 seconds. Add this to our time it took to rise and we get a total time of 4.95 seconds.

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• At what clock time(s) will the speed of the ball be 5 meters / second?

answer/question/discussion: ->->->->->->->->->->->-> : I think that this will only happen once because on the way down it will have a speed of -5m/s.

vf – v0 / a = (5m/s – 15m/s) / -10m/s^2 = 1 second.

it starts at 15 m/s on the way up, and decreases to 0 m/s at the top, so it has to be 5 m/s at some point on the way up

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• At what clock time(s) will the ball be 20 meters above the ground?

answer/question/discussion: ->->->->->->->->->->->-> : This will happen twice. Once on the way up and again on the way down.

We started at 12 meters above the ground so we need to calculate how long it will take for the ball to be at 20 meters or increase 8 meters.

8m/ 15m/s = .533 seconds to reach 20 meters going up. We calculated earlier that we reach our maximum height of 34.5 meters. So to figure when we reach the 20 meter mark again we need to calculate how long it will take for the ball to fall 14.5 meters. 14.5m / 10m/s^2 = 1.45 seconds. This added to our 1.5 seconds that it takes to get to maximum height gives us a total of 2.95 seconds to reach the 20 meter mark on the way down.

UP: .533 seconds

Down: 2.95 seconds.

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• How high will it be at the end of the sixth second?

answer/question/discussion: ->->->->->->->->->->->-> :

If we are at 34.5 meters off the ground at the end of 1.5 seconds and decrease this at 10m/s^2 for the next 4.5 seconds will result with the ball on the ground. Over the 4.5 seconds we will travel a total of 45 meters. We would hit the ground at approx 5 seconds.

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30 Minutes

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