Asked by Jordan
A referee tosses a coin straight up with a velocity of 5.25 m/s, how high does it go above its point of release? (Hint: How fast is it moving at the maximum height?)
Answers
Answered by
Damon
The vertical velocity is 0 at the top
v = Vi - 9.81 t
0 = 5.25 - 9.81 t
t = .535 second to the top
h = 0 + Vi t - 4.9 t^2
h = 5.25 (.535) - 4.9 (.535^2)
h = 1.40 meter
alternate and faster way:
(1/2) m v^2 = m g h
[ conservation of energy ]
h = v^2/2g = 5.25^2/19.6 = 1.40 meter
v = Vi - 9.81 t
0 = 5.25 - 9.81 t
t = .535 second to the top
h = 0 + Vi t - 4.9 t^2
h = 5.25 (.535) - 4.9 (.535^2)
h = 1.40 meter
alternate and faster way:
(1/2) m v^2 = m g h
[ conservation of energy ]
h = v^2/2g = 5.25^2/19.6 = 1.40 meter
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