#$&*
Phy 121
Your 'cq_1_08.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** **
A ball is tossed upward at 15 meters / second from a height of 12 meters above the ground. Assume a uniform downward acceleration of 10 m/s^2 (an approximation within 2% of the 9.8 m/s^2 acceleration of gravity).
• How high does it rise and how long does it take to get to its highest point?
answer/question/discussion: ->->->->->->->->->->->-> :
vf = v0 + a ‘dt
0m/s = 15m/s + (-10m/s^2)* ‘dt
10m/s^2 * ‘dt = 15m/s
‘dt = 1.5s Clock time at highest point.
vAve = (vf + v0)/2
vAve = (0m/s + 15m/s)/2
vAve = 7.5m/s
vAve = ‘ds/’dt
‘ds = vAve * ‘dt
‘ds = 7.5m/s * 1.5s
‘ds = 11.25m Max height above starting position.
#$&*
• 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: ->->->->->->->->->->->-> :
‘ds = -11.25m + -12m
‘ds = -23.25m
‘ds = v0 ‘dt + 0.5 a ‘dt/2
v0 = 0
0.5 a ‘dt^2 = ‘ds
‘dt^2 = 2* ‘ds/a
‘dt = +-sqrt(2* ‘ds/a)
‘dt = +-sqrt(2* -23.25m/-10m/s^2)
‘dt(downward) = +-2.16s
‘dt(downward) = 2.16s
‘dt(upward) = 1.5s
‘dt = ‘dt(upward) + ‘dt(downward)
‘dt = 1.5s + 2.16s
‘dt = 3.66s Clock time when the ball strikes the ground.
vAve = ‘ds/’dt
vAve = -23.25m/2.16s
vAve = -10.76m/s
vAve = (vf + v0)/2
-10.76m/s = (vf + 0m/s)/2
vf = 2 * -10.76m/s
vf = -21.52 m/s Velocity when the ball strikes the ground.
#$&*
• At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
vf = v0 + a ‘dt
5m/s = 15m/s + (-10m/s^2)* ‘dt
10m/s^2 * ‘dt = 15m/s - 5m/s
10m/s^2 * ‘dt = 10m/s
‘dt = 1s
#$&*
• 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: ->->->->->->->->->->->-> :
vAve(upward) = 7.5m/s
‘ds = 20m - 12m
‘ds = 8m
vAve = ‘ds/’dt
7.5m/s = 8m/’dt
‘dt = 8m/7.5m/s
‘dt = 1.07s Clock time at 20m on way up.
vAve(downward) = -10.76m/s
-23.25m - ‘ds = -20m
‘ds = -23.25m + 20m
‘ds = -3.25m
vAve = ‘ds/’dt
-10.76m/s = -3.25m/‘dt(downward)
‘dt(downward) = -3.25m/-10.76m/s
‘dt(downward) = 0.30s
‘dt(upward) = 1.5s
‘dt = ‘dt(upward) + ‘dt(downward)
‘dt = 1.5s + 0.30s
‘dt = 1.80s Clock time at 20m on way down.
The ball strikes the ground at 3.66s so the ball will be on the ground after 6s.
#$&*
** **
30 minutes
** **
&#Very good responses. Let me know if you have questions. &#
@&