Question

D-11. 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 2 seconds of the release of
second object will be
(1) 4.9 m
(2) 9.8 m
(3) 19.6 m
(4) 24.5 m​

Answer 1

Answer:

24.5 m

Explanation:

After 2 seconds of release of object 2.

The object 1 take 3 seconds and object 2 take 2 seconds.

So the distance travelled by first object in 3 seconds = ut + 1/2 at^2 = 44.1 m

and distance travelled by second object in 2 seconds = ut +1/2 at^2 = 19.6 m

So the distance between them = 44.1 - 19.6 = 24.5 m

Answer 2

Answer:

The distance between the two objects is 24.5 m.

(4) is correct option.

Explanation:

Given that,

Time for first object = 1 sec

Total time t = 1+2= 3 sec

Time for second object = 2 sec

We need to calculate the distance of the first object

Using equation of motion

$s_{1} = ut+\dfrac{1}{2}gt^2$

Put the value in the equation

$s_{1}=0+\dfrac{1}{2}\times9.8\times3^2$

$s_{1}=44.1\ m$

We need to calculate the distance of the second object

Using equation of motion

$s_{2} = ut+\dfrac{1}{2}gt^2$

Put the value in the equation

$s_{2}=0+\dfrac{1}{2}\times9.8\times2^2$

$s_{2}=19.6\ m$

We need to calculate the distance between the two objects

$D=s_{1}-s_{2}$

$D=44.1-19.6=24.5\ m$

Hence, The distance between the two objects is 24.5 m.