Q. 194.2( 13 Votes )

# 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?

Answer :

let h_{1}=150m and h_{2}=100m

Then, the difference in height h=(150-100)m=50 m

Distance travelled by first body in 2s is given as,

Distance=ut + at^{2}

Now, if the Initial speed(u) = 0(when the object was at rest)

Ans, the time, t =2s

So, by substituting the values, we get,

S = 0 +g× (2)^{2}

S = 0 +g× 4

S = 0 +× 9.8 × 4

S = 19.6 m

So, After 2s, the height of the first object, h_{1}=150-19.6

So, After 2s height of the second object, h_{2}=100-19.6

Thus, after 2s difference of height= h_{1} - h_{2} (150-19.6) - (100-19.6) =50 m

So, after 2s difference height will be 50 m = initial difference in height

Thus, the difference in height of the object will remain same because the height does not vary with time.

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