phy 121
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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: ->->->->->->->->->->->-> :
v=v0 + at
0 = 15 m/s + (-10m/s^2)t
-15 m/s = (-10 m/s^2)t
1.5s = t It reaches highest point at 1.5 sec.
y= y0 +v0t + 1/2 at^2
y= 12 m + 15m/s(1.5s) + 1/2(-10m/s^2)(1.5s)^2
y= 12 m + 22.5 m + -11.25m
y=23.25m
#$&*
How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground?
answer/question/discussion: ->->->->->->->->->->->-> :
When it gets back to where it was tossed it will be going -15 m/s. Then it has an additional 12 meters to travel so if it is accelerating at -10 m/s^2 then
it will be going a little over -25 m/s when it hits the ground. It should accomplish this in around 3 seconds.
#$&*
At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :The ball will be going this speed just after it is tossed due to the downward acceleration of gravity. Then once it reaches the top height of 23.25 m
it will stop and soon after the 1.5 second market be accelerating downward at this negative speed.
#$&*
At what clock time(s) will the ball be 20 meters above the ground?
How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
Between one and one and a half seconds, it will be this distance above the ground. At 6 sec. the ball should be on the ground.
#$&*
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Solution
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