A body is thrown vertically up with a velocity u.if it passes 3 points A,B,C in its upward journey with the velocity U/2, U/3, U/4 respectively. The ratio of separation AB and BC is?

Step 1: Given

Initial velocity of the body = u.

Velocity at point A, $u_A=\frac{U}{2}$

Velocity at point B, $u_B=\frac{U}{3}$

Velocity at point C, $u_C=\frac{U}{4}$

Step 2: Formula used and calculation

To find distance AB, use third equation of motion, $v^2=u^2+2as$

Here, $u=\frac{U}{2},a=-g,v=\frac{U}{3}$

$\therefore {\left(\frac{U}{3}\right)}^2={\left(\frac{U}{2}\right)}^2+2(-g)s_{AB}$

$\therefore s_{AB}=\frac{5U^2}{72g}$

Similarly to find distance BC, use third equation of motion

Here, $u=\frac{U}{3},a=-g,v=\frac{U}{4}$

$\therefore {\left(\frac{U}{4}\right)}^2={\left(\frac{U}{3}\right)}^2+2(-g)s_{BC}$

$\therefore s_{BC}=\frac{7U^2}{288g}$

Now,

$\frac{s_{AB}}{S_{BC}}=\frac{\frac{5U^2}{72g}}{\frac{7U^2}{288g}}$

$\frac{s_{AB}}{S_{BC}}=\frac{20}{7}$