#$&*
Phy 121
Your 'cq_1_08.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_08.2_labelMessages **
A ball is tossed upward at 15 meters / second from a height of 12 meters above the ground. Assume a uniform downward acceleration of 10 m/s^2 (an approximation within 2% of the 9.8 m/s^2 acceleration of gravity).
How high does it rise and how long does it take to get to its highest point?
answer/question/discussion: ->->->->->->->->->->->-> sion:
We can reason that the ball will reach its highest point when it reaches a velocity of 0m/s which will happen after 1.5 seconds.
At this time, its vAVe will be 15m/s/2 = 7.5m/s
`ds = 7.5m/s * 1.5s = 11.25 meters we add this to the original height of 12 m to get a height of 23.25m
#$&*
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: ->->->->->->->->->->->-> sion:
from its height of 23.25 m it will fall 10m/s/s. If we now assume the apex is the new start point we have three values v0, a and `ds. Solve using an equation
vf^2 = v0^2 + 2 a `ds
vf^2 =0^2 + 2a 10m/s^2 *23.25m
vf^2 = 465m^2/s^2 Take `sqrt of each side
vf = 21.6m/s (approximately)
vAVe = 21.6m/s/2 = 10.8 m/s
`dt = `ds/vAve = 23.25m/10.8m/s = 2
2 seconds from apex means 3.5 seconds after intitial toss.
#$&*
@&
Good.
Note that you would get these results by assuming v0 = 15 m/s, a = -9.8 m/s^2 and `ds = -12 meters. It may be convenient to do so, especially if you've already figured out the clock time and position at the apex, but it isn't necessary to break this part of the question into two separate phases of motion.
*@
At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> sion:
At 1 second, the speed will be 5m/s. 15m/s -10m/s = 5m/s
And maybe at 2 seconds, because it is 1.5 seconds to the apex and another .5s of falling at 10m/s/s. Although I believe with speed, direction is imortant so maybe this is -5m/s and doesn't count.
#$&*
@&
Good.
Note that you would get these results by assuming v0 = 15 m/s, a = -9.8 m/s^2 and `ds = -12 meters. It may be convenient to do so, especially if you've already figured out the clock time and position at the apex, but it isn't necessary to break this part of the question into two separate phases of motion.
*@
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: ->->->->->->->->->->->-> sion:
vf^2 = v0^2 + 2 a `ds
vf^2 =(15m)^2 + 2*-10m/s^2 *8m
vf^2 = 225m^2/s^2 - 160m^2/s^2
vf^2 = 65m^2/s^2 Take `sqrt of each side
vf = 8.1m/s (approximately)
vAVe = 8.1m/s/2 = 4.05 m/s
`dt = `ds/vAve = 8m/4.05m/s = 2s Approximately.
That doesn't seem to be right since it should be at max height in 1.5 seconds. I think I have some general concepts down but something seems off.
On the way back down, if we assume we are at 0m/s we can use the formula
vf^2 = v0^2 + 2 a `ds
vf^2 =v0^2 + 2*10m/s^2 *3.25m Since 3.25 from the apex is 20 meters (on the way down)
vf^2 = 20m/s *3.35m
vf^2 = 65m^2/s^2 Take `sqrt of each side
vf = 8.1m/s (approximately)
vAVe = 8.1m/s/2 = 4.05 m/s
`dt = `ds/vAve = 3.25/4.05m/s = .8s Approximately.
We add this to the 1.5 seconds it takes to get to the apex and we get 2.3 s
Lastly, after 6 seconds how high will the ball be.
15m/s - 10m/s^2 *6s = 15m/s -60m/s = -45m/s
vAVe = (15m/s + - 45m/s = -30m/s
-30m/s *6 secons is -180 meters.
The ball will be on the ground.
#$&*
*#&!
&#Your work looks good. See my notes. Let me know if you have any questions. &#