Question

physics a flower pot falls from a windowsill 25.0m above the sidewalk. how much time does a passerby on the sidewalk below have to move out of the way before the flowerpot hits the ground? i need help!!! i suck at physics

Answer 1

$v_f^2 = v_0^2 + 2*a*d$
Since the flower pot is dropped, it has an initial velocity of zero. Also the flower pot accelerates due to gravity.

so

$v_f = \sqrt{2*g*d}=\sqrt{2*9.81*25.0}$