Asked by Ashley
A. A baseball is seen to pass upward by a window 28.1 m above the street with a vertical speed of 13.7 m/s. If the ball was thrown from the street, what was its initial speed?
B. What altitude does it reach?
C. How long before the ball passed the window was it thrown?
D. How long after the baseball passed the window does it reach the street again?
B. What altitude does it reach?
C. How long before the ball passed the window was it thrown?
D. How long after the baseball passed the window does it reach the street again?
Answers
Answered by
drwls
A. Total energy per mass = Vo^2/2
= (13.7)^2/2 + g*28.1
Solve for initial speed, Vo
B. Vo^2/2 = g*Hmax
Solve for max altitude, Hmax
c. (Velocity change)/g = time going up
= (Vo - 13.7)/g
d. (time spent going up after passing window) + (fall time)
= 13.7/g + Vo/g
= (13.7)^2/2 + g*28.1
Solve for initial speed, Vo
B. Vo^2/2 = g*Hmax
Solve for max altitude, Hmax
c. (Velocity change)/g = time going up
= (Vo - 13.7)/g
d. (time spent going up after passing window) + (fall time)
= 13.7/g + Vo/g
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