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Question

A stone is dropped from the top of a building. When it crosses a point 5m below the top, another stone starts to fall from a point 25m below the top, both stones reach the bottom of building simultaneously. The height of the building is: Takeg=10ms2


A

45m

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B

35m

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C

25m

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D

50m

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Solution

The correct option is A

45m


Step 1: Given data assumptions.

Height covered by the first stone=5m

Height covered by the second stone=25m

Acceleration due to gravity, g=10ms2

Step 2: Finding The height of the building.

Formula Used:

S=ut+12gt2

Where, S is distance, u initial velocity, and t is time

Since,

The velocity of the first stone at 5m below top, u1=2gh=2×10×5=10ms

Now,

Apply a formula for the first stone which is shown in the figure we get,

20+h=u1t+12gt2

20+h=10t+12gt2 …..i

Apply a formula for the second stone which is shown in the figure we get,

h=ut+12gt2

h=12gt2 u=0 …..ii

Substitute equation ii in i we get.

20+12gt2=10t+12gt2

t=2s ….iii

Now, again substitute equation iii in equation ii we get.

h=12×10×22

h=20m

Hence, the height of the building from the drawn figure=h+25=20+25=45m

Therefore, option A is correct.


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