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: ->->->->->->->->->->->-> :
Where does velocity equal 0
15m/s – (10m/s^2 (x sec)) = 0m/s
15m/s = (10m/s^2 (x sec))
X = 15m/s / 10m/s^2
X=1.5s It takes 1.5 seconds to reach the highest point
vAve = 15m/s+0m/s / 2 = 7.5m/s
7.5m/s * 1.5s = 11.25m
Highest point: 11.25m from the starting point, or 23.25m off the ground
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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: ->->->->->->->->->->->-> :
Fourth equation of motion: vf^2 = v0^2 +2a (ds)
Vf^2 = (15m/s)^2 + 2(-10m/s)(-12m)
Vf^2 = 225m^2/s^2 +240m^2/s^2
Vf^2 = 465m^2/s^2
Vf = 21.6m/s final velocity
Third equation: ds = v0(dt)+.5a(dt^2)
-12 = 15m/s(dt)+.5(-10m/s^2)(dt^2) solved via the quadratic equation
Dt = 3.66 seconds after the toss, the object reaches the ground
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• At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
Calculate for +/- 5m/s because you are taking into account direction (up and down)
15m/s –(10m/s^2 (x s)) = 5m/s
-10m/s^2 (x s) = -10m/s
x = 1 second
15m/s –(10m/s^2 (x s)) = -5m/s
-10m/s^2 (x s) = -20m/s
X = 2 seconds
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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: ->->->->->->->->->->->-> :
With the present calculations, 0 m is the take off point and -12 m is the ground, you have to calculate when it is at 8m, to find when it is really at 20 meters.
A= -10m/s^2
V= -10t+c1 c1 = 15m/s v = -10t+15
S = -10/2 t^2 +c1t+c2 c2 = 0 s = -5t^2 +15t
-5t^2+15t = 8
-5t^2+15t-8 = 0 quadratic equation
T = .694 s and 2.31 s
6 seconds: 15m/s – (10m/s^2(6s)) = 15m/s-60m/s = -45m/s
-45m/s *6 seconds = -270 m
It is already on the ground because the height is much lower than the allowed -12m
To verify: we found that the object reached the ground at 3.66 seconds
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30 minutes
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