Phy 231
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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: ->->->->->->->->->->->-> :
v0=15m/s
vf=0m/s
a= -10m/s^2
so `dv= -15m/s we find that `dt= 1.5s
vAve=15/2=7.5m/s*1.5s= 11.25m+12m= 23.25m
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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: ->->->->->->->->->->->-> :
v0=0m/s
A= -10m/s^2
`ds= 23.25m
Vf^2=0^2+2(-10)(23.25)= -21.6m/s
10.8m/s= 23.25m/`dt so `dt for this part= 2.15s….. but we must add the first 1.5s
2.15s+1.5= 3.65s
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• At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
5m/s=15m/s+(-10m/s^2)dt= 1s and
When the initial velocity is 0m/s and the vf= -5m/s
-5m/s=-10m/s^2(`dt)= 0.5s but we must add the initial time so .5+1.5= 2s
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• At what clock time(s) will the ball be 20 meters above the ground?
20 m = 12 m + 15 m/s 'dt – 5 m/s^2 'dt^2
0 = -5 m/s^2 * 'dt^2 + 15 m/s – 8
use the quadratic.( - 15 +- sqrt(225 – (4)(-5)(-8))]/(2)(-5) =
( -15 +- squareroot( 225 – 160)/-10 = -15 +- sqaureroot(65)/ -10 = (-15 +- 8.06)/-10 = -23.06/ -10 & -6.92/ -10 = 2.31s and .69s
• How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
For the ball to go all the way up into the air and back down to the grounf it only took 3.65s, therefore we know that the ball is on the ground after 6s. So the distance off of the gorund is 0m.
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35 minutes
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