Question

Two particle move in a uniform gravitational field with an acceleration $g$. At the initial moment the particles were located over a tower at one point and moved with velocities $v_1=3m/s$ and $v_2=4m/s$ horizontally opposite directions. Find the distance between the particles at the moment when their velocity vectors become mutually perpendicular.

Answer 1

The horizontal component of velocity of both particles remains same as there is no force in the horizontal direction.

Thus after $t$ sec.

The vertical component of velocity of the particles.

$v_{y1}=v_{y2}=0\cdot t+\frac{1}{2}g{t}^2$

and their horizontal components are $v_{x1}=3$

$v_{x2}=4$

Let the angle made by the velocity vectors with the horizontal be ${\theta }_{1}and{\theta }_2$.

Thus,

$tan{\theta }_1=\frac{\frac{1}{2}g{t}^2}{3}tan{\theta }_2=\frac{\frac{1}{2}g{t}^2}{4}$

Given that

${\theta }_1+{\theta }_2={90}^{0}tan({\theta }_1+{\theta }_2)=\infty$

Substituting values of

$tan{\theta }_{1}andtan{\theta }_2$

$\frac{\frac{\frac{1}{2}g{t}^2}{4}+\frac{\frac{1}{2}g{t}^2}{3}}{1-\frac{\frac{1}{4}{g}^2{t}^4}{12}}=\infty$

Solving above, we get

$1-\frac{1}{4}\times \frac{g^2{t}^4}{12}=0$

Or

$t^4=\frac{48}{{\left(9.8\right)}^2}t=\sqrt[4]{4\frac{48}{{\left(9.8\right)}^2}}=0.841secs$

Relative horizontal velocity V of the particles

$V=3-(-4)=7m/sec$

Distance D between two particles when their velocity vectors are at right angles to each other

$D=7\times 0.841=5.887metres$