Question

$A \: body \: is \: thrown \: vertically \\ upward \: with \: the \: initial \: velocity \: \\ v1. Another \: body \: B \: is \: dropped \\ from \: a \: height \: h. \: Find \: how \: the \\ distance \: x \: between \: the \: bodies \: \\ depends \: on \: the \: time \: t \: of \: the \\ bodies \: begin \: to \: move \: \\ simultaneously \\ \\ (1)x = h - v1t \\ (2)x = (h - v)t \\ (3)x = h - \frac{v1}{t} \\ (4)x = \frac{h}{t} - v1 \: 7$

#Answer the question with steps#

Answer 1

answer is 1st option..

hope I made it clear.......
☺☺☺☺

Answer 2

Hey there =_=

:cry: Brainliest must await for another answer

Anyways 0_0

Consider the two Velocity Vectors :

(I) For the Particle thrown with velocity v from Ground

$\vec{v_1} = (v - gt) \hat{j}$

(II) For the Particle dropped from height h

$\vec{v_2} = (- gt) \hat{j}$

$=> \vec{v_{2/1}} =\vec{v_2} - \vec{v_1} = (-v) \hat{j}$

Integrating with respect to time, and having the initial positions( with frame of reference = particle 1 ),

$\vec{x_{2/1}} = (h - vt) \hat{j}$

And so, the relative position of The Particle 2, with respect to 1, is given by ( h - vt )