Asked by Aly
a baseball is popped straight up. the ball leaves the bat moving at 37.8 m/s. at what time is the ball 20 m above the ground?
Answers
Answered by
Henry
See your previous post: Sun,12-1-13,10:34 PM.
Answered by
Henry
h = Vo*t + 0.5g*t^2 = 20
37.8t _ 4.9t^2 = 20
-4.9t^2 + 37.8t - 20 = 0
Use Quadratic Formula and get:
t = 0.571, and 7.14 s.
So, when the ball is rising; it reaches
20 m in 0.571 s. When it is falling, it
reaches 20 m in 7.14 s.
37.8t _ 4.9t^2 = 20
-4.9t^2 + 37.8t - 20 = 0
Use Quadratic Formula and get:
t = 0.571, and 7.14 s.
So, when the ball is rising; it reaches
20 m in 0.571 s. When it is falling, it
reaches 20 m in 7.14 s.
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