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Phy 121
Your 'cq_1_08.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_08.2_labelMessages **
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: ->->->->->->->->->->->-> :
Well we know that when it gets to the highest point, v = 0 m/s. Therefore we say that going upward, v0 is equal to 15 m/s and vf = 0 m/s. The acceleration is -10 m/s^2, so we can use the acceleration formula to find `dt.
a = `dv / `dt, therefore `dt = `dv / a = (0 m/s - 15 m/s) / -10 m/s^2 = -15 m/s / -10 m/s^2 = 1.5 s
The vAve for the interval of the 1.5 s is equal to the (vf + v0) / 2. vAve = (0 m/s + 15 m/s) / 2 = 7.5 m/s
Now that we know vAve we can solve for how high it rises by using the vAve formula.
vAve = `ds / `dt
`ds = `dt * vAve = 1.5 s * 7.5 m/s = 11.25 m
Therefore it rises to a maximum of 11.25 m in 1.5 s.
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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: ->->->->->->->->->->->-> :
Well if we want to know how long it took after the initial toss, then we take the acceleration, the initial velocity, and the change in distance from the highest point (which we know must be -11.25 m + -12 m) Therefore we can find how long it took, and then how fast it was traveling when it hits the ground.
I am going to find the final velocity first, so I am going to use the 3rd uniformly accelerated equation using the second half of the path traveled (not half I guess, but when velocity reaches 0 m/s at its highest point). vf^2 = v0^2 + 2 a `ds = (0 cm/s)^2 + 2 * -10 m/s^2 * -23.25 m = 465 m^2/s^2
Now to just find vf we have to take the square root of both sides. vf = sqrt(465 m^2/s^2) = +-21.6 m/s, since we are counting as the direction of velocity to be upward, this 21.6 m/s must be negative since we are going against the set direction. Therefore the final velocity is -21.6 m/s.
Now we need to find vAve. vAve = (15 m/s + -21.6 m/s) / 2 = -3.3 m/s
So now we use this information to solve for `dt.
`dt = `ds / vAve = -23.25 m / -3.3 m/s = 7.0 s. So now we add this to the first interval time of 1.5 seconds and get 8.5 s for the time for the ball to strike the ground after the initial toss.
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At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
I would assume that we need to take a = `dv / `dt to find this.
The initial velocity is 15 m/s so now we are going to plug in 5 m/s in for vf and use the a = -10 m/s^2 to find `dt. `dt = (vf - v0) / a = (5 m/s - 15 m/s) / -10 m/s^2 = -10 m/s / -10 m/s^2 = 1 s
Since speed cannot be negative and velocity can, I would assume that the moment when velocity was -5 m/s would count too. `dt = (vf - 0) / a = (-5 m/s - 15 m/s) / -10 m/s^2 = -20 m/s / -10 m/s^2 = 2 s
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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 have no idea how to figure this problem out. I feel like this problem isn't too difficult to find, but I just don't know where to begin. These questions are really difficult for me to solve for. If I could see another example very similar to this one, I could probably reason out how to do it.
I also have no idea how to figure out how high it will be at the end of the 6th second.
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