Asked by Jordan
A baseball is popped straight up into the air and has a hang-time of 6.25 s. Determine the height to which the ball rises before it reaches its peak.
Answers
Answered by
Aman singla
v=u at
0=u (-9.8) (25/8)
u=30.5m/s
s=ut 1/2 (at0…5)
s=30.5(25/8) 1/2(-9.8)(25/8)0…5
s=(1525/16) (-3062.5/64)
s=1525/16 - 30625/640
61000-30625/640
30375/640
47.46m
0=u (-9.8) (25/8)
u=30.5m/s
s=ut 1/2 (at0…5)
s=30.5(25/8) 1/2(-9.8)(25/8)0…5
s=(1525/16) (-3062.5/64)
s=1525/16 - 30625/640
61000-30625/640
30375/640
47.46m
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