a ball is projected vertically upward with velocity 9.8 m/s . what is avg velocity of ball during ascent
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Average velocity is the total displacement / total time
So the time taken would be 1 sec to come to 0
And displacement would be
9.8(1)-4.9(1*1)
4.9 and hence
The average velocity would be 4.9 m/sec
So the time taken would be 1 sec to come to 0
And displacement would be
9.8(1)-4.9(1*1)
4.9 and hence
The average velocity would be 4.9 m/sec
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GIVEN:
initial velocity= u= 9.8 m/s
Acceleration due to gravity= -9.8m/s2
Let h be the height travelled during time of ascent.
At maximum height , Final velocity will be zero , V=0
V^2-U^2=2as
V^2-U^2=2gh
0-(9.8)^2=2×(-9.8)h
h= 9.8x9.8/2*9.8
h=4.9m
Time of ascent = u/g
=9.8/9.8
=1 sec
Therefore average velocity= displacement/ time
=4.9/1
=4.9 m/s
Therefore,avg velocity of ball during ascent 4.9m/s
initial velocity= u= 9.8 m/s
Acceleration due to gravity= -9.8m/s2
Let h be the height travelled during time of ascent.
At maximum height , Final velocity will be zero , V=0
V^2-U^2=2as
V^2-U^2=2gh
0-(9.8)^2=2×(-9.8)h
h= 9.8x9.8/2*9.8
h=4.9m
Time of ascent = u/g
=9.8/9.8
=1 sec
Therefore average velocity= displacement/ time
=4.9/1
=4.9 m/s
Therefore,avg velocity of ball during ascent 4.9m/s
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