Question

What is meant by freefall? A ball is dropped from the roof of a building. It takes 10 seconds to reach the ground. Find the height of the building. \[\left( {g = 9.8m/{s^2}} \right)\]

Answer 1

Hint: We can simply use the second equation of motion i.e. $s = ut + \dfrac{1}{2}a{t^2}$ where the initial speed is zero , acceleration is acceleration due to gravity and time is 10 seconds.

Complete step by step answer:A free falling object is an object which is falling from a height only under the influence of gravity (no other force is acting on the object). So, any object falling solely under the influence of gravity is said to be in a state of freefall. While considering an object in a freefall, air resistance is assumed to be zero and objects are falling with an acceleration of \[g = 9.8m/{s^2}\] .
A ball is dropped from the top of a building, so it will experience free fall. Also, the vertical distance covered by the ball will be equal to the height of the building because the ball can only cover that distance if its thrown from top and hits the ground.
Let’s see the information given in the question
Time taken (t)= 10s
Initial velocity(u)= 0 m/s
Acceleration (a)= \[g = 9.8m/{s^2}\]
Height/Vertical distance (s)=?
Using the second equation of motion, we can write
$
  s = ut + \dfrac{1}{2}a{t^2} \\
\Rightarrow s = 0m/s \times 10s + \dfrac{1}{2} \times 9.8m/{s^2} \times {\left( {10s} \right)^2} \\
\Rightarrow s = 490m \\
 $
So, the answer is the building is 490 meters tall.

Note:We can also use the first and third equations of motion to find the distance covered by the ball as follows.
$
  v = u + at \\
  v = 98m/s\left( {9.8m/{s^2} \times 10s} \right) \\
 $
And substituting this value in the third equation of motion
$
  {v^2} - {u^2} = 2as \\
\Rightarrow {98^2} - 0 = 2 \times 9.8 \times s \\
   \Rightarrow s = 490m \\
 $