#$&*
phy 201
Your 'cq_1_08.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** CQ_1_08.2_labelMessages **
A ball is tossed upward at 15 meters / second from a height of 12 meters above the ground. Assume a uniform downward acceleration of 10 m/s^2 (an approximation within 2% of the 9.8 m/s^2 acceleration of gravity).
How high does it rise and how long does it take to get to its highest point? answer/question/discussion:
v0=+15m/s up
a=-10m/s^2 down
vf=0
0=15-10*dt
-15=-10*dt
dt=1.5s
vAve=15/2=7.5m/s
ds=7.5*1.5=11.25m#$&*
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:
v0=+15m/s up
a=-10m/s^2 down
ds-12m
vf^2=15^2+2(-10)-12
vf^2=225+240
vf=sqrt465
vf=-22m/s
-22=15-10*dt
-37=-10dt
dt=3.7s #$&*
At what clock time(s) will the speed of the ball be 5 meters / second? answer/question/discussion:
v0=+15m/s up
vf=+5m/s
a=-10m/s^2
5=15-10dt
-10=-10dt
dt=1s #$&*
@&
Good, but there is another answer as well. Speed is 5 m/s when velocity is 5 m/s or -5 m/s.
*@
At what clock time(s) will the ball be 20 meters above the ground?
v0=+15m/s up
ds=8m
a=-10m/s^2
vf^2=15^2+2(-10)8
225-160
vf=sqrt65
vf=8.1
8=15+8.1/2*dt
8=11.5dt
dt=.7s
How high will it be at the end of the sixth second?answer/question/discussion:
The ball hit the ground at 3.7 seconds
v0=+15m/s up
a -10ms^2
dt=6s
vf=15-10*6
vf=-45m
.. ( or on the ground) #$&*
http://vhcc2.vhcc.edu/dsmith/forms/common_questions/ph1_cq_I_08_2.htm
&#Good responses. See my notes and let me know if you have questions. &#