Phy 231
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.
** **
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?
v0 = 15 m/s
p1 = 12 m
a = -10 m/s^2
so, since vf would be 0 m/s, then vAve for the interval between initial throw and highest distance would be (0 m/s + 15 m/s)/2 = 7.5 m/s
if a = -10 m/s^2, and a = dv/dt, and dv = (0 m/s - 15 m/s), then (-15 m/s)/dt = -10 m/s^2, so dt = 1.5 sec
and then vAve *dt = ds, so (7.5 m/s)(1.5 s) = 11.25 m higher than the starting point (or 23.25 m from the ground)
*
How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground?
well, from the highest point to the ground, ds = -23.25 m
and we know that it took the ball 1.5 s to reach the highest point, so since
ds = v0(dt) + .5a(dt)^2
(-23.25 m) = (0 m/s)(dt) + .5(-10 m/s^2)(dt)^2
0 = -5 m/s^2 (dt)^2 + 0 - 23.25 m
5 m/s^2 (dt)^2 = - 23.25 m
-dt = +- sqrt(4.65)
since seconds can only be positive dt = 2.16 s
So it took 2.16 seconds to fall to the ground after reaching the highest point, or (2.16 + 1.5) seconds = 3.66 seconds after throwing it to reach the ground
and since vAve = ds/dt, and ds for the entire throw is -12 m (counting the 12 meter starting point as the origin), so vAve = -12m/3.66 sec = -3.3 m/s, and since v0 is 15 m/s, and vAve = (v0 + vf)/2, then -3.3 m/s = (15 m/s + vf)/2 and we can solve for vf = -21.6 m/s
*
At what clock time(s) will the speed of the ball be 5 meters / second?
Since vAve = (v0 + vf)/2, then vAve = (15 m/s + 5 m/s)/2 = 10 m/s, and
a = dv/dt, so (5 m/s - 15 m/s)/dt = -10 m/s, and dt = 1 second
The speed will be 5 m/s at two different instants, the second being when the velocity is -5 m/s
*
At what clock time(s) will the ball be 20 meters above the ground?
20 meters above ground = a displacement of + 8 (12+8 m = 20 m), and
ds = v0(dt) + .5a(dt)^2
8 m = (15 m/s)(dt) + .5(-10 m/s^2)(dt)^2, so
0 = -5 m/s^2 (dt)^2 + 15 m/s (dt) - 8 m
and dt = 1.2 seconds
*
How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> scussion:
Well, it will have hit the ground long before second six, but if there were a hole in the ground,
a = dv/dt
-10 m/s^2 = dv/6 sec
dv = -60 m/s
so since v0 = 15 m/s, then vf = -60 m/s + 15 m/s = -45 m/s
and vAve = (15 m/s + -45 m/s)/2 = -15 m/s
and (-15m/s)(6 sec) = -90 m below the starting point, or -78 m below the ground
** **
10-15 min
** **
&#Your work looks good. See my notes. Let me know if you have any questions. &#