course Phy 201
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:
Since we have acceleration of gravity at 10m/s^2 and we are throwing the ball upward we will have a -10m/s^2 acceleration of
gravity. We start out with a velocity of 25m/s and if the the acceleration is -10m/s^2 then after 1 second the velocity will
be 15m/s. 25m/s-10m/s=15m/s
What will be its velocity at the end of two seconds?
answer/question/discussion:
Same applies on this one only we will be subtracting from the position at 1 second. 15m/s-10m/s=5m/s
During the first two seconds, what therefore is its average velocity?
answer/question/discussion:
We add v0 and vf and divde by 2. 25m/s/5m/s=30m/s/2=15m/s
How far does it therefore rise in the first two seconds?
answer/question/discussion:
We take the average velocity and multiply it by the time interval which gives us 30m. 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:
At three seconds it will be traveling at -5m/s. After .5 seconds the ball is at its maximum height at 0m/s and starts its way
back to the ground. So after one second it is traveling at -5m/s
At what instant does the ball reach its maximum height, and how high has it risen by that instant?
answer/question/discussion:
The ball has reach its highest distance at 0m/s. At this point the ball is neither going up or going down.
The distance the ball has traveled is 31.25m. We get this by taking (v0+vf)/2=vAve. 25m/s+0m/s=25m/s/2=12.5m/s*2.5s=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:
Since we already know at 3secs we are traveling at -5m/s and acceleratin of gravity is 10m/s^2 we will be traveling at -15m/s
after 4seconds.
To find the distance we take the velocity at its highest peak which is 0m/s and the velocity at 4seconds which is -15m/s. We
add these and divide by 2 to get vAve which is 7.5m/s. We take vAve*`ds= 7.5m/s*1.5s=11.25m. We then subtract this from are
highest peak of 31.25m. 31.25m-11.25m=20m. The distance from the ground after 4s is 20m
How high will it be at the end of the sixth second?
answer/question/discussion:
Below the earths surface somewhere.
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