Question

A body A is thrown vertically upward with the initial velocity ${v_1}$. Another body B is dropped from a height h. Find how the distance x between the bodies depends on the time t if the body begins to move simultaneously.
A. $x = (h - {v_1})t$
B. $x = h - {v_1}t$
C. $x = h - \dfrac{{{v_1}}}{t}$
D. $x = \dfrac{h}{t} - {v_1}$

Answer 1

Hint: In order to solve this question, we will find the distance covered by both the bodies in the same interval of time, and then we will subtract the addition of their distance from net height h which will be the distance $x$ between the bodies.

Formula used:
Equation of motion is,
$S = ut + \dfrac{1}{2}a{t^2}$
where,
S is the distance covered by the body
u is the initial velocity of the body
a is the acceleration of the body and when body motion is under effect of gravity then $a = g$ called acceleration due to gravity.
t is the time taken by the body.

Complete step by step solution:
For body A which is projected with initial velocity $u = {v_1},a = - g$ upwards, suppose it covers a distance of $S = {H_A}$ in time t so, using equation of motion $S = ut + \dfrac{1}{2}a{t^2}$ we get,
${H_A} = {v_1}t - \dfrac{1}{2}g{t^2} \to (i)$
For Body B which is dropped under the effect of gravity so, $u - 0,a = g$ downwards, suppose it covers a distance of $S = {H_B}$ in time t, so we have;
${H_B} = \dfrac{1}{2}g{t^2} \to (ii)$
So, distance between the bodies A and B will be
$x = h - ({H_A} + {H_B})$
On putting the values we get,
$\Rightarrow x = h - ({v_1}t) \\
\therefore x = h - {v_1}t \\ $
Hence, the correct answer is option B.

Note: While solving such kinematics problems, never confuse with sign convention of gravity and distances always make sure for a falling body freely gravity is taken as negative while for a body projecting upward gravity is taken as negative while other quantities in newton’s equation of motion taken as positive in each cases.