Asked by Daniela
A ball player catches a ball 3.33s after throwing it vertically upward.
a) with what speed did he throw it?
b) what height did it reach?
a) with what speed did he throw it?
b) what height did it reach?
Answers
Answered by
Damon
time in air = 3.33
so
time ascending = 3.33/2 = 1.665
v = 0 at top = Vi - 9.81 (1.665)
so
Vi = 16.3 m/s
average speed up = 16.3/2
time up = 1.665
so distance up = 13.6 meters
so
time ascending = 3.33/2 = 1.665
v = 0 at top = Vi - 9.81 (1.665)
so
Vi = 16.3 m/s
average speed up = 16.3/2
time up = 1.665
so distance up = 13.6 meters
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