Question

A penny falls from a windowsill which is 25.0 m above the sidewalk. How much time does a passerby on the sidewalk below have to move out of the way before the penny hits the ground?

Answer 1

x = vt +1/2(g)t^2
x is height, v is initial velocity, g is acceleration due to gravity, t is time
v is 0
x = 1/2(g)t^2
solve for t
t = (2x/g)^(1/2)
t = (50/9.81)(1/2)
t = 2.258 seconds

Answer 2

second to last line should be
t = (50/9.81)^(1/2)