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  • An object is dropped from rest at a height of 150 m and simultaneously another object is dropped from rest at a height 100 m. What is the difference in their heights after 2 s if both the objects drop with same accelerations?

An object is dropped from rest at a height of 150 m and simultaneously another object is dropped from rest at a height 100 m. What is the difference in their heights after 2 s if both the objects drop with same accelerations? How does the difference in heights vary with time?

Answers (1)

As the objects are dropped from rest their initial velocity is zero.

The acceleration will be acceleration due to gravity, which will be equal to 10 ms-2.

The height of the particle is written as:

D = H-S

Where H is initial height and S is displacement.

If we write heights of both the particles by using second equation of motion: s = ut + \frac{1}{2}at^2

D_{1}=H_{1}- \left ( ut + \frac{1}{2}at^{2} \right )

D_{2}=H_{2}- \left ( ut + \frac{1}{2}at^{2} \right )

Hence,

D_{1}-D_{2}=H_{1}-H_{2}=50m

The difference in height at any time is coming out to be 50 m.

So even, after two seconds, the difference will be 50 m and it will not vary with time.

Posted by

infoexpert21

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