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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:
(15m/s)/(10m/s^2)=1.5s
Ds=v0*dt+.05adt
Ds=(15m/s)*(1.5s)+.05*(-10m/s^2)(1.5s)^2
Ds=21.38m
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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:
vf^2=v0^2+2a*ds
ds=21.38m
vf^2=((0m/s)^2)+2(10m/s^2)(21.38m)
vf=20.68m/s
20.68m/s=0m/s+(10m/s^2)*dt
Dt=2.07s
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Remember that the ball started from a height of 12 meters.
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0m/s=15m/s+(-10m/s^2)*dt
Dt=0s
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The solution to the preceding equation is `dt = 1.5 s, not `dt = 0 s.
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At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion:
vf=v0+a*dt
dt=(vf-v0)/a
dt=(5m/s)/(10m/s^2)
dt=.5s
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The ball starts with an upward velocity of 15 m/s.
After .5 s its speed will be 10 meters / second.
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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:
I am not sure how to calculate this, I am not sure how to reason it out or have been able to successfully manipulate an equation of uniformly accelerated motion.
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The ball starts at height 12 m.
So it has to rise another 8 m to get to 20 m.
Its initial velocity is 15 m/s, its acceleration is -10 m/s^2.
Knowing `ds, v0 and a, I expect that you'll be able to solve the problem.
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Very good, but you do have some oversights and there are a couple of places where I'm not completely sure what you've done.
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