Question

The height of the Empire State Building is 318.00 meters. If a stone is dropped from the top of the building, what is the stone's velocity just before it strikes the ground?

Answer 1

Assuming there is no wind resistance and the stone has no initial velocity, use the equation
\[V_f^2 =V_i^2 +2a(x_f-x_i)\]
Or you should have some other equation that looks like it.
Vf is the final velocity we are looking for.
Vi is the initial velocity, which is zero
a is the acceleration due to gravity (9.81)
xi is the starting spot for the stone (0)
xf is the final location for the stone (318)
plug it all in and solve for Vf
\[v_f^2 = (0)^2 +2(9.81)(318-0)\]
\[V_f^2 =6239.16\]
\[V_f =79 m/s\]

Answer 2

Use
\[ d=\frac{ 1}{ 2 }g t^{2}\]
solve for time
t=\[t=\sqrt{2d/g}\]
t=8.0518 sec

Answer 3

v = sqrt(2 x g x h)
v = sqrt(2 x 9.81 x 318)
v = sqrt(6239.16)
v = 78.98 m/s

right???