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?
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: ->->->->->->->->->->->-> :
How high does it rise and how long does it take to get to its highest point?
v0 = +15m/s
vf = 0
a = -10 m/s^2
0 = 15m/s - 15m/s + -10 m/s^2 (`dt)
-15m/s / -10m/s^2 = -10m/s^2 / -10m/s^2 (`dt)
`dt = 1.5 seconds
To find the height:
`ds = (vf + v0) / 2 * `dt
`ds = (0 + 15m/s) / 2 * 1.5s
`ds = 7.5m/s * 1.5s
`ds = 11.25 meters
Add this to 12 which is the height above ground from where the ball was thrown (12 + 11.25 meters = 23.25m)
It takes 1.5 seconds to reach its maximum height at 23.25 meters.
How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground?
The `ds would be 23.25m from the peak to end + the 12m to the ground, the total is 35.25m.
v0 = 23.25m/1.5s = 15.5m/s
vf = 0
a = -10 m/s^2
0 = 15.5m/s + -10 m/s^2 (`dt)
15.5m/s / -10m/s^2 = -10m/s^2 / -10m/s^2 (`dt)
`dt = 1.55s
It takes 1.55 seconds to reach the ground from the peak. Add this to the 1.5 seconds it takes to reach the peak, the total time for the toss is 3.1 seconds.
The velocity when it hits the ground is 22.7m/s (35.25m/1.55s = 22.7m/s).
At what clock time(s) will the speed of the ball be 5 meters / second?
v0 = +15m/s
vf = 5m/s
a = -10 m/s^2
vf = v0 + a`dt
5m/s = 15m/s -10m/s^2 (`dt)
5m/s 15m/s = -10m/s^2(`dt)
-10m/s / -10m/s^2 = -10m/s^2 / -10m/s^2 (`dt)
`dt = 1 second
At what clock time(s) will the ball be 20 meters above the ground?
`ds = 20m
v0 = +15m/s
vf = (peak) 23.25m / 1.5s = 15.5m/s
`ds = (vf + v0) / 2 * `dt
20m = (15.5m/s + 15m/s) /2 * `dt
20m / 15.25m/s = 15.25m/s / 15.25m/s * `dt
`dt = 1.3 seconds
How high will it be at the end of the sixth second?
v0 = +15m/s
vf = ?
a = -10 m/s^2
`dt = 6s
Vf = v0 + a`dt
Vf = 15m/s 10 m/s^2 (6s)
Vf = 15m/s 60m/s
Vf = -45m/s
`ds = (vf + v0) / 2 * `dt
`ds = (-45m/s + 15m/s) / 2 * 6s
`ds = -15m/s * 6s
`ds = -90m
If the peak is at 23.25m, from the start point back to the starting point its 46.5m. After 6 seconds it would be -43.5m from the starting point.
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