Question 62: Motion in 1D

Question

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 :

Accepted Answer

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)
$$h = {1 \over 2}g{t^2}$$
$$ = {1 \over 2} 	imes 10 	imes {2^2}$$
h = 20 m
The height of the building $$ = 25 + 20$$ = 45 m

Answer

50 m

Answer

25 m

Answer

45 m

Answer

35 m

Motion in 1D

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 :

Choose the correct answer: