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PHY121
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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: ->->->->->->->->->->->-> :
After one second, the ball would have a velocity of 15m/sec. v0 of 25m/sec - acc of gravity 10m/s^2
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What will be its velocity at the end of two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
After two seconds, the velocity would be 5m/sec.
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During the first two seconds, what therefore is its average velocity?
answer/question/discussion: ->->->->->->->->->->->-> :
vAve = 15m/sec + 5m/sec= 10m/sec
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How far does it therefore rise in the first two seconds?
answer/question/discussion: ->->->->->->->->->->->-> :
The ball rises 20m in the first two seconds. (10m/sec *2sec)
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What will be its velocity at the end of a additional second, and at the end of one more additional second?
answer/question/discussion: ->->->->->->->->->->->-> :
After 3 seconds>>>-5m/sec
After 4 seconds>>>-15m/sec
The ball is returning downward.
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At what instant does the ball reach its maximum height, and how high has it risen by that instant?
answer/question/discussion: ->->->->->->->->->->->-> :
The ball reaches max height between two and three seconds. It is at max height somewhere close to 20m.
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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: ->->->->->->->->->->->-> :
vAve= (5+15+(-5)+(-15))/4=0 At the end of the 4th second, the ball is on the ground.
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How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
0m
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*#&!*#&!
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