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 are,
(A) $3.41\sec ,57m$
(B) $4.31\sec ,57m$
(C) $1.34\sec ,57m$
(D) $3.14\sec ,75m$

Answer 1

Hint: We solve this question by using the second equation of motion to find the time of free fall. We first get an equation for distance travelled by body in time $t$. We then find an equation for distance travelled by body in time $t-1$. Subtracting both equations, we get the time of free fall as well as the distance of free fall.
Formula used: second equation of motion is equal to $s=ut+{\displaystyle \frac{1}{2}}a{t}^2$
Here, Distance is represented by $s$
Initial velocity is represented by $u$
Time is represented by $t$
Acceleration is represented by $a$

Complete step by step answer:
Since the body is falling freely the initial velocity will be zero
The second equation of motion for free fall is equal to $s={\displaystyle \frac{1}{2}}a{t}^2$
Finding the distance travelled by the body in time $t$
Let us consider the distance travelled by body in time $t$ as $s_{total}$
$s_{total}={\displaystyle \frac{1}{2}}g{t}^2$ .......(1)
Here, we take acceleration as acceleration due to gravity which is represented by $g$ because in a free fall the body is accelerated only due to gravity.
Now finding the distance travelled by body in time $t-1\sec$
Let us consider the distance travelled by body in time $t-1\sec$ as $s_{total-1}$
 $s_{total-1}={\displaystyle \frac{1}{2}}g(t-1{)}^2$........(2)
Subtracting (2) from (1) we get the distance travelled in the last one second.
$s_{total}-s_{total-1}={\displaystyle \frac{1}{2}}g(2t-1)$
From the question the distance travelled by body in the last one second is equal to $s/2$
Substituting this in the above equation we get
${\displaystyle \frac{s}{2}}={\displaystyle \frac{1}{2}}g(2t-1)$......(3)
From equation (1) $s_{total}={\displaystyle \frac{1}{2}}g{t}^2$
Equation (3) now becomes ${\displaystyle \frac{g{t}^2}{4}}={\displaystyle \frac{1}{2}}g(2t-1)$
${\displaystyle \frac{t^2}{4}}=t-{\displaystyle \frac{1}{2}}$
Solving for t we get
$0=t^2-4t+2$
Solving the polynomial equation, we get,
$t=0.59,t=3.41\sec$
From the question the body travels only half the distance in the last one second this means that the total time must be greater than one sec. Thus, time of free fall is equal to $t=3.41\sec$
We height if free fall is nothing but the total distance travelled by the body. From equation (1)
$s_{total}={\displaystyle \frac{1}{2}}g{t}^2$
Substituting the value of $t=3.41\sec$
$s_{total}={\displaystyle \frac{1}{2}}\times 9.8\times {3.41}^2$
$s_{total}=57m$

Hence option (A) $3.41\sec ,57m$ is the correct answer.

Note: It is to be remembered that in a free fall of a body the initial velocity is always zero unless the question mentions something else. Also, the only acceleration acting on the body will be the acceleration due to gravity unless the question mentions any additional information. One must remember these points to not make a mistake.