Asked by jona
I'm not sure how to solve this..a baseball is popped straight up into the air and has a hang-time of 6.25s. Determine the height to which the ball rises before it reaches its peak.
Answers
Answered by
Damon
time to top = 6.25/2 = 3.125
v = Vi - 9.8 t
0 = Vi - 9.8 t
Vi = 9.8*3.125 = 30.625
so
h = 0 + Vi t - (9.8)/2 t^2
h = 30.625 * 3.125 - 4.9 * (3.125)^2
= 47.85 meters
v = Vi - 9.8 t
0 = Vi - 9.8 t
Vi = 9.8*3.125 = 30.625
so
h = 0 + Vi t - (9.8)/2 t^2
h = 30.625 * 3.125 - 4.9 * (3.125)^2
= 47.85 meters
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