Question

A stone falls freely from rest, and the total distance covered by it in the last second of its motion equals the distance covered by it in the first three seconds of the motion. Find the time for which the stone remains in the air and the total height from where the stone is dropped.

Answer 1

Hint :If it is a free fall, then it means that the acceleration is acting in the downwards direction with the magnitude equal to $ g = 9.8\,m\,{s^{ - 2}} $ . We shall apply the second speed equation and find a relation between the distance covered in three seconds. Since it is a free fall the initial velocity has to be zero. We shall also the formula of the distance travelled in the nth second and find the answer after substituting the known values.
The speed equations are
 $ v = u + at $
 $ s = ut + \dfrac{1}{2}a{t^2} $
 $ 2as = {v^2} - {u^2} $
Where u is the initial velocity, v is the final velocity, s is the distance covered, t is the time taken and a is the acceleration.
The distance travelled by an object in nth second is $ {s_n} = u + \dfrac{{(2n - 1)a}}{2} $ where u is the initial velocity of the motion and a is the acceleration.

Complete Step By Step Answer:
Using the second speed equation,
 $ s = ut + \dfrac{1}{2}a{t^2} $ where u is the initial velocity, s is the distance covered, t is the time taken and a is the acceleration.
Substituting the known values we get,
 $ s = 0t + \dfrac{1}{2}g \times {3^2} $
 $ \Rightarrow s = 4.5g $
Now the distance travelled in the last second is the same as the distance we just calculated.
Substituting in the formula we get,
 $ 4.5g = \dfrac{{(2n - 1)g}}{2} $ where n is the number of seconds the stone is in the air
 $ \Rightarrow 2n - 1 = 9 $
 $ \Rightarrow n = 5 $
So, the stone remains in the air for five seconds.
Again, using the second speed equation,
 $ s = ut + \dfrac{1}{2}a{t^2} $
 $ \Rightarrow s = 0 + \dfrac{1}{2} \times 9.8 \times {5^2} $
 $ \Rightarrow s = 122.5\,m $
So, the distance covered or the total height is 122.5 m.

Note :
We must very carefully note which values have been given to us. It is often very easy to visualize the situation given in such questions. We must also carefully notice the sign of these quantities. In this particular question we used the second speed equation since the question did not say a thing about the final velocity of the stone. So, we need to choose the equations accordingly.