Asked by stephanie
A penny is dropped from the observation deck of the Empire State building (370 m above ground). With what velocity does it strike the ground? Ignore air resistance.
Answers
Answered by
DrBob222
distance = 1/2 gt^2 Solve for time it takes for the penny to reach the ground.
Then distance = 1/2(Vinitial + Vfinal)t.
Vinitial is zero. You know distance and t, solve for Vfinal.
Then distance = 1/2(Vinitial + Vfinal)t.
Vinitial is zero. You know distance and t, solve for Vfinal.
Answered by
Anonymous
12
Answered by
Anonymous
85.2
Answered by
Tesfalem
d=1/2gt^2
t^2=2d/g
t^2=2(370)/9.8
t^2=75.51
t=8.6 second
v2=v1+gt
v2=0+(9.8m/s^2)(8.6s)
v2=84.3m/s
t^2=2d/g
t^2=2(370)/9.8
t^2=75.51
t=8.6 second
v2=v1+gt
v2=0+(9.8m/s^2)(8.6s)
v2=84.3m/s
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