Question

A tourist drops a penny off the Empire State Building. The penny hits the ground 12 seconds later. What is the velocity of the penny when it hits the ground?

Answer 1

Answer:

117.6 m/s downward

Explanation:

The penny is moving with uniform accelerated motion (free fall motion), so we can find its vertical velocity at time t with the following suvat equation:

v = u + at

where

v is the velocity at time t

u is the initial velocity

a is the acceleration

The penny starts at rest, so

u = 0

Also, the acceleration is the acceleration of gravity,

$a=g=-9.8 m/s^2$

where we have chosen upward as positive direction, so the acceleration is negative, since it is downward.

Substituting t = 12 s, we find the velocity of the penny when it hits the ground:

$v=0 + (-9.8)(12)=-117.6 m/s$

downward, since the sign is negative.