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 begin to move simultaneously.

• A. $x=(h-v_1)t$
• B. $x=h-v_{1}t$
• C. $x=h-\frac{v_1}{t}$
• D. $x=\frac{h}{t}-v_1$

Answer 1

The correct option is B $x=h-v_{1}t$

Let $h_1$ and $h_2$ be the distance travelled by the body $A$ and $B$ respectively in time $t$, and $x$ be the distance between the bodies.
Then, we have
$h_1=v_{1}t-\frac{g{t}^2}{2}$ and $h_2=\frac{g{t}^2}{2}$....(i)
So, we have distance between the bodies
$x=h-(h_1+h_2)$
From (i), $h_1+h_2=v_{1}t$
Hence, we have
$x=h-v_{1}t$