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course Phy 121
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:
Velocity at the highest point is 0. Initial velocity is 15m/s. So using the equation v^2 = vf^2 +2adt, and solving it for dt, we get a height of 11.25 meters + the 12 meters it was above the ground, so we get a total height of 23.25 meters.
Using the equation v = v0 + at and solving for time, and plugging in the numbers from above, we end up using the equation t = v-v0/a. Since the velocity at the top is 0, we get 15m/s/ 10m/s^s = 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:
It takes the ball the same amount of time to return to the 12 m above the ground, so it takes it 3 seconds, plus 1.2 seconds to cover the remaining 12m, for a total of 4.2 seconds.
It takes less than 1.2 seconds to travel that last 12 meters. The object's speed is increasing, and when it first gets back to its original level it's already moving faster that 10 m/s.
To find the velocity at time 4.2 seconds, we use the equation vf = v0 + at. Plugging in the numbers, we get vf = 15m/s + (10m/s^2)(4.2s) = 57 m/s.
Some quantities are positive, some negative. In your solution they are all positive.
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• At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion:
I’m assuming the question is asking for at what time the ball will reach 5m/s after leaving the throwers hand. the equation vf = v0 +at, and plugging in 5m/s for vf and 15m/s for v0, as well as 10 m/s^2, we get 1 second after leaving the hand.
Good use of the equation, but you can reason that out very directly using the definition of acceleration.
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• 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:
The ball is traveling at an average velocity of 15m/s +0m/s/2, so 7.5 m/s. At 20 meters above the ground, it must be 8 meters above the throwers hand. Meaning the ball travels 8 metes with 7.5 m/s velocity, it takes just over 1 second.
Using the equation vf = v0 +at, and plugging in the numbers from above, we get vf = 15m/s +10ms*6. The final velocity at the end of 6 seconds is 75 m/s. Moving that fast, only being 23.3 meters above the ground, the ball will have already struck the ground and have a velocity of 0.
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