Question

An object is released from some height exactly after one second another object is released from the same height. The distance between the two objects exactly after two seconds of the release object will be:

A. $4.9\,m$

B. $9.8\,m$

C. $19.6\,m$

D. $24.5\,m$

Answer 1

Hint: Here we have to use the second equation of motion. The second equation of motions gives the relation between time, velocity, acceleration and speed. The value of g should be taken as negative while the object is going upside.

Complete step by step solution:

Let ${S_1}$ ​ be the distance travelled by the first object released. Similarly ${S_2}$ be the distance travelled by the second object released for the second object. After $2$ seconds of release of object $2$ , object $1$ takes $3$ seconds and object $2$ takes $2$ seconds. Due to different release times of two objects, the distance swept by both objects are different. If any two objects are released at the same time then the distance covered by both objects will be the same irrespective of their mass.

The acceleration on the objects is equal to the acceleration due to gravity since no other force is acting on the objects that can create the acceleration. Its value is $9.8$ metres per second square. The formula for distance travelled from height in $t$ seconds is calculated as: 

$ut + \dfrac{1}{2}a{t^2}$

Here, the initial velocity ($u$) is zero as initially they are at rest position.

so , ${S_1} = $ the distance travelled by first object in $3$ seconds $ = ut + \dfrac{1}{2}a{t^2}= 0+ \dfrac{1}{2} \times (9.8){(3}^2) = 44.1\,m$

And ${S_2} = $ distance travelled by second object in $2$ seconds $ = ut + \dfrac{1}{2}a{t^2}= 0+\dfrac{1}{2} \times (9.8)({2}^2) = 19.6\,m$

The distance between the two objects exactly after two seconds of the release object is given by ${S_1} - {S_2}$.

$\Rightarrow {S_1} - {S_2} = 44.1 - 19.6$

$\Rightarrow {S_1} - {S_2} = 24.5\,m $

The distance between the two objects exactly after two seconds of the release object $ = 24.5\,m$

Hence, Option (D) is the correct answer.

Note: Due to different release time of two objects , distance swept by both objects are different. If any two objects are released at the same time then the distance covered by both objects will be the same irrespective of their mass.