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 bodies 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 A $x=h-v_{1}t$

Let the distance covered by body $A$ be $h_1$
$\Rightarrow h_1=v_{1}t-\frac{1}{2}g{t}^2$
and, let the distance travelled by the body $B$ be $h_2$
$\Rightarrow h_2=0+\frac{1}{2}g{t}^2$
Given, the distance between the bodies is $x=h-(h_1+h_2)$
$\because h_1+h_2=v_{1}t$
$\Rightarrow x=h-v_{1}t$
Hence, the distance $x$ between the bodies depends on the time $t$ if the bodies begin to move simultaneously as $x=h-v_{1}t$