PHY 201
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Seed 8.2
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?
How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground?
At what clock time(s) will the speed of the ball be 5 meters / second?
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:
After 1.5 seconds the ball has reached its highest point. The vAve after 1.5 seconds is 7.5 m/s. 7.5 * 1.5 = 11.25 meters
This is the distance traveled. Add this to 12 meters which is where it was thrown from and the heighth of the ball above ground is 23.25 meters.
After 3 seconds of being thrown the ball is at a height of 12 meters with velocity 15 m/s.
'ds = 12
v0 = 15
'dt = ?
a = 10 m/s/s
'dt = 'ds/[(vF + v0)/2]
a = (vF - v0)/'dt
a = (vF - v0)/[ds/[(vF + v0)/2]]
a * 'ds/[(vF + v0)/2) = vF - v0
a * 2 * 'ds/(vF + v0) = vF - v0
a * 2 * 'ds = vF^2 - v0^2
vF^2 = 2a 'ds + v0^2
= 240 + 225
= 465
vF = 21.6 m/s
Therefore the 'dv = 6.6
So 10 = 6.6/'dt
'dt = .66 seconds
So that means that the ball hit the ground at a speed of 21.6 m/s 3.66 seconds after being thrown up
Times the ball will be 5 m/s:
a = 10 m/s/s
v0 = 15 m/s
vF = 5 m/s
'dv = 10 m/s
'dt?
10 = 10/'dt
'dt = 1 second
So the ball will be traveling at 5 m/s 1 second after being thrown up and also 2 seconds after being thrown up.
When will the ball be 20 meters above ground:
The balls highest point is 23.25 meters.
a = -10 m/s/s
v0 = 15 m/s
'ds = 8 m
vF^2 = 2(-10)(8) + 15^2
= -160 + 225
vF = 8.06 m/s
'dv = 6.94
10 = 6.94/'dt
'dt = .694 seconds
After it reaches its highest point it will need to travel 3.25 meters to reach 20 again.
a = 10
v0 = 0
'ds = 3.25
vF = 8.06 (sqt of 65)
10 = 8.06/'dt
'dt = .806
Add this to 1.5 = 1.856 seconds
So the ball will be 20 meters above ground .694 seconds after being tossed up and also 1.856 seconds after being tossed up
How high will it be after 6 seconds:
It has already hit the ground
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45 minutes
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Very good work, but check out the link (revision is again optional and probably unnecessary, but you can be the judge).
&#At least part of your solution does not agree with the solution and comments given at the link below. You should view the solution at that link and self-critique as indicated there.
Solution
This link also expands on these topics and alerts you to many of the common errors made by students in the first part of this course. &#