Question

Two particles are simultaneously projected in opposite direction horizontally from a given point in space where gravity g is uniform.If $u_{1}and{u}_2$ be their initial speeds, then the time t after which their velocities are mutually perpendicular is given by

• A. $\frac{\sqrt{u_1{u}_2}}{g}$
• B. $\frac{\sqrt{u_1^2+u_2^2}}{g}$
• C. $\frac{\sqrt{u_1(u_1+u_2)}}{g}$
• D. $\frac{\sqrt{u_2(u_1+u_2)}}{g}$

Answer 1

Answer: (A)

$\frac{\sqrt{u_1{u}_2}}{g}$

$v_1=u_1\hat{i}-gt\hat{j}$

$v_2=-u_2\hat{i}-gt\hat{j}$

$for{v}_1\perp v_2$

${\overline{v}}_1.{\overline{v}}_2=0$

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

$-u_1{u}_2+g^2{t}^2=0$

$g^2{t}^2=u_1{u}_2$

$t=\frac{\sqrt{u_1{u}_2}}{g}$