A man of $$50\,kg$$ mass is standing in a gravity free space at a height of $$10\,m$$ above the floor. He throws a stone of $$0.5\,kg$$ mass downwards with a speed $$2\,m{s^{ - 1}}.$$ When the stone reaches the floor, the distance of the man above the floor will be
A.
$$9.9\,m$$
B.
$$10.1\,m$$
C.
$$10\,m$$
D.
$$20\,m$$
Answer :
$$10.1\,m$$
Solution :
As, $$mr = {\text{constant}}$$
$$\eqalign{
& {m_1}{r_1} = {m_2}{r_2} \Rightarrow {r_2} = \frac{{{m_1}{r_1}}}{{{m_2}}} \cr
& = \frac{{0.5 \times 10}}{{50}} = 0.1 \cr} $$
So, distance travelled by the man will be $$10 + 0.1 = 10.1\,m$$
Releted MCQ Question on Basic Physics >> Impulse
Releted Question 1
The figure shows the position-time $$\left( {x - t} \right)$$ graph of one dimensional motion of a body of mass $$0.4kg.$$ The magnitude of each impulse is
A rigid ball of mass $$m$$ strikes a rigid wall at $${60^ \circ }$$ and gets reflected without loss of speed as shown in the figure. The value of impulse imparted by the wall on the ball will be
The force $$F$$ acting on a particle of mass $$m$$ is indicated by the force-time graph shown below. The change in momentum of the particle over the time interval from $$0$$ to $$8 s$$ is
A man of $$50\,kg$$ mass is standing in a gravity free space at a height of $$10\,m$$ above the floor. He throws a stone of $$0.5\,kg$$ mass downwards with a speed $$2\,m{s^{ - 1}}.$$ When the stone reaches the floor, the distance of the man above the floor will be