First slide

Projection Under uniform Acceleration

Question

Two particles move in a uniform gravitational field with an acceleration g. At the initial moment the particles were located at one point and move with velocities $v_{1} =$ 3.0 m/s and $v_{2} =$ 4.0 m/s horizontally in opposite directions. Find the distance between the particles at the moment when their velocity vectors become mutually perpendicular.

Moderate

A

5 m
B

$7 \sqrt{3} m$
C

$\frac{7 \sqrt{3}}{5} m$
D

$\frac{7}{2} m$

Solution

Let the particle move perpendicular to each other at time t.
Here $(4 \overset{⏞}{i} - gt \overset{⏞}{j}) . (- 3 \overset{⏞}{j} - gt \overset{⏞}{j}) = 0$
$\Rightarrow - 12 + g^{2} t^{2} = 0$
$\Rightarrow t = \sqrt{(\frac{12}{100})} \Rightarrow t = \frac{\sqrt{3}}{5}$
Hence distance,
d = $\sqrt{{(4 t + 3 t)}^{2} + ((h - \frac{1}{2} {gt}^{2}) - {(h - \frac{1}{2} {gt}^{2}))}^{2}}$
= $7 t$
$= \frac{7 \sqrt{3}}{5} m$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App