All Problems

Kinematics

Problem 1.11

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

Reveal Answer
l=(v1+v2)v1v2/g=2.5 ml=\left(v_{1}+v_{2}\right) \sqrt{v_{1} v_{2}} / g=2.5 \mathrm{~m}