Asked by Anonymous
A baseball is seen to pass upward by a window 29 m above the street with a vertical speed of 15 m/s. The ball was thrown from the street.After how many more seconds does it reach the street again?
Answers
Answered by
MathMate
let
u=initial velocity = 15 m/s
g=acceleration due to gravity = 9.8 m/s/s
S=-29 m
Using the equation
S=ut+(1/2)(-g)t²
the only unknown is the time t.
You will get two solutions, the negative one is when it was thrown from the street, and the other is the solution sought, and should be over 4 seconds.
u=initial velocity = 15 m/s
g=acceleration due to gravity = 9.8 m/s/s
S=-29 m
Using the equation
S=ut+(1/2)(-g)t²
the only unknown is the time t.
You will get two solutions, the negative one is when it was thrown from the street, and the other is the solution sought, and should be over 4 seconds.
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