Question

Two balls are thrown horizontally from the top of a
building with speed u, and u, respectively in
opposite directions. The separation between two
balls when they are moving perpendicular to each
other, is (ans. √u1u2/g×(u1+u2)​

Answer 1

first find out velocity vector of each ball.

velocity vector of first ball after time t, $\vec{v_1}=u_1\hat{i}-gt\hat{j}$

similarly, velocity vector of 2nd ball after time t, $\vec{v_2}=-u_2\hat{i}-gt\hat{j}$

if they are moving perpendicular,

$\vec{v_1}.\vec{v_2}=0$

or, $(u_1\hat{i}-gt\hat{j}).(-u_2\hat{i}-gt\hat{j})=0$

or, $-u_1.u2+g^2t^2=0$

or, t = $\frac{\sqrt{u_1u_2}}{g}$...(1)

let seperation between two ball is h,

h = $u_1t +u_2t$

= $u_1\frac{\sqrt{u_1u_2}}{g}+u_2\frac{\sqrt{u_1u_2}}{g}$ [ from eq (1). ]

= $(u_1+u_2)\frac{\sqrt{u_1u_2}}{g}$

hence, seperation between the ball when they are perpendicular to each other is $(u_1+u_2)\frac{\sqrt{u_1u_2}}{g}$

Answer 2

Answer:see the attachment

Explanation: