JEE MAIN - Physics (2021 - 25th February Evening Shift - No. 18)
A stone is dropped from the top of a building. When it crosses a point 5 m below the top, another stone starts to fall from a point 25 m below the top. Both stones reach the bottom of building simultaneously. The height of the building is :
50 m
25 m
45 m
35 m
Explanation
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For particle (1)
$$20 + h = 10t + {1 \over 2}g{t^2}$$ ...... (i)
For particle (2)
$$h = {1 \over 2}g{t^2}$$ ..... (ii)
put equation (ii) in equation (i)
$$20 + {1 \over 2}g{t^2} = 10t + {1 \over 2}g{t^2}$$
t = 2 sec.
Put in equation (ii)
$$h = {1 \over 2}g{t^2}$$
$$ = {1 \over 2} \times 10 \times {2^2}$$
h = 20 m
The height of the building $$ = 25 + 20$$ = 45 m
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