#$&*
course Phy 121
A ball is tossed upward with an initial velocity of 25 meters / second. Assume that the acceleration of gravity is 10 m/s^2 downward. What will be the velocity of the ball after one second?
answer/question/discussion:
1 sec*-10m/s^2= -10m/s
-10m/s+25m/s= 15m/s
#$&*
What will be its velocity at the end of two seconds?
answer/question/discussion:
2*-10m/s^2= -20m/s
-20m/s+25m/s=5m/s
#$&*
During the first two seconds, what therefore is its average velocity?
answer/question/discussion:
(25m/s+5m/s)/2=15m/s
#$&*
How far does it therefore rise in the first two seconds?
answer/question/discussion:
15m/s * 2s =30m
#$&*
What will be its velocity at the end of a additional second, and at the end of one more additional second?
answer/question/discussion:
3s*-10m/s^2= -30m/s
-30m/s+25m/s=-5m/s
4s*-10m/s^2=-40m/s
-40m/s+25m/s=-15m/s
#$&*
At what instant does the ball reach its maximum height, and how high has it risen by that instant?
answer/question/discussion:
after 2 sec the ball is at 5m/s and at 3s it is at -5m/s, therefore it reached it reached 0m/s at the midpoint of that interval which is at 2.5s.
ds=v0*dt+.5(a)(dt^2)
ds=25m/s*2.5s+.5(-10m/s)(2.5s)^2
ds=31.25m
#$&*
What is its average velocity for the first four seconds, and how high is it at the end of the fourth second?
answer/question/discussion:
-10m/s*4s=-40m/s+25m/s= -15m/s
(15m/s+5m/s-5m/s-15m/s)/4 =0m/s
The ball has traveled 2*31.25m=62.5m
#$&*
How high will it be at the end of the sixth second?
answer/question/discussion:
it will be at 0m, because it will have come back to the ground and stoped.
@&
... unless there's a hole in the ground ...
*@
#$&*
" ""
&#Your work looks very good. Let me know if you have any questions. &#