Question

The observation deck of a tall skyscraper is 370m above the street. Determine the time required for a penny to free fall from the deck to the street below.
A. $8.69s$
B. $7.69s$
C. $5.69s$
D. $8.6s$

Answer 1

Hint: The time taken by the penny to reach the street below can be found by using the equations of motion for freely falling objects. Firstly, we need to find the final velocity of the penny as it reaches the ground. Then, using the velocity we can find the time it took the penny to fall.
Formula used:
${v^2} - {u^2} = 2gh$
$v - u = gt$

Complete answer:
In the question, they’ve given that a penny is dropped from a deck which is at a height of 370m from the ground.
Since the penny is dropped i.e. freely falling object it does not have any initial velocity. And its acceleration will be that of acceleration due to gravity. Therefore, it will be moving at constant acceleration.
The velocity of the penny on reaching the street below will be given by
${v^2} - {u^2} = 2gh$
Where,
v is the final velocity of the penny
u is the initial velocity of the penny
g is the acceleration due to gravity
h is the height of the building
Substituting the values in the equation, we have
$\eqalign{
  & {v^2} - {u^2} = 2gh \cr
  & \Rightarrow {v^2} - 0 = 2 \times 9.8 \times 370 \cr
  & \Rightarrow {v^2} = 7252 \cr
  & \Rightarrow v = 85.16m{s^{ - 1}} \cr} $
On reaching the ground the penny will have a velocity of 85.16m/s.
The first equation of motion is given by
$v - u = gt$
Substituting the values,
$\eqalign{
  & v - u = gt \cr
  & \Rightarrow 85.16 - 0 = 9.8 \times t \cr
  & \Rightarrow t = \dfrac{{85.16}}{{9.8}} = 8.69s \cr
  & \therefore t = 8.69s \cr} $

So, the correct answer is “Option A”.

Note:
The acceleration of any freely falling object will always be equal to the acceleration due to gravity. And the equations of motion will only apply for objects moving with constant acceleration. And the equation we’ve used above are the equations of motion for freely falling bodies.