Question

If a body travels half its total path in the last second of its fall from rest. The time and height of its fall will respectively
(A) $ 0.59{\text{s, 57m}} $
(B) $ 3.41{\text{s, 57m}} $
(C) $ 5.9{\text{s, 5}}{\text{.7m}} $
(D) $ 5.9{\text{s, 34}}{\text{.1m}} $

Answer 1

Hint : To solve this question, we need to use the second kinematic equation of motion. There are two conditions given in the question, with the help of which two equations will be generated, solving which we will get the two unknowns.

Formula Used: The formula which is used to solve this question is given by
 $\Rightarrow s = ut + \dfrac{1}{2}a{t^2} $ , here $ s $ is the displacement covered in time $ t $ with an initial velocity of $ u $ and acceleration of $ a $ .

Complete step by step answer
Let the time and the height of the fall of the body be $ T $ and $ H $ respectively.
From the second equation of motion we have
 $\Rightarrow s = ut + \dfrac{1}{2}a{t^2} $ ............................(1)
Since the object is falling from rest, so $ u = 0 $ . Also, it is covering a height of $ H $ in time $ T $ . Therefore we have
 $\Rightarrow H = \dfrac{1}{2}g{T^2} $ ............................(2)
Now, it is given that the body covers half the total path in the last second. We have the total path of the body equal to the height of its path. So the half of the total path becomes $ \dfrac{H}{2} $ . This distance is covered in the last second, which means that the rest half of the total path is covered in $ t = \left( {T - 1} \right)s $ . Substituting $ s = \dfrac{H}{2} $ , and $ t = \left( {T - 1} \right)s $ in (1) we get
 $\Rightarrow \dfrac{H}{2} = \dfrac{1}{2}g{\left( {T - 1} \right)^2} $
 $ \Rightarrow H = g{\left( {T - 1} \right)^2} $ ............................(3)
Equating (2) and (3)
 $\Rightarrow \dfrac{1}{2}g{T^2} = g{\left( {T - 1} \right)^2} $
 $ \Rightarrow {T^2} = 2{\left( {T - 1} \right)^2} $
Expanding the term on the RHS
 $\Rightarrow {T^2} = 2{T^2} - 4T + 2 $
 $ \Rightarrow {T^2} - 4T + 2 = 0 $
On solving we get
 $\Rightarrow T = \left( {2 \pm \sqrt 2 } \right)s $
We know that $ \sqrt 2 \approx 1.41 $ . So we have
 $\Rightarrow T = 0.59s,{\text{ }}3.41s $
As half of the total path is covered by the body in $ 1s $ , so the total time of fall must be more than $ 1s $ . Thus, the value $ T = 0.59{\text{s}} $ is rejected. Therefore, the time of fall is $ T = 3.41{\text{s}} $ . From (2) the height of fall is given as
 $\Rightarrow H = \dfrac{1}{2} \times 9.8 \times {3.41^2} $
On solving we get
 $\Rightarrow H = 56.98m $
 $ \Rightarrow H \approx 57m $
So the correct answer will be option B.

Note
We could not use the information for the half path length which is covered in the last second. This is because that would require the velocity of the body at the midpoint of the path which needs to be separately calculated. So we used the relation for the first half of the path length covered to save our time and efforts.