UPSEE Solved Model Paper 7

Given: Initial velocity of the body is u At point A : $v_1$ = $\frac{u}{2}$ At point B : $v_2$ = $\frac{u}{3}$ At point C : $v_3$ = $\frac{u}{4}$ As the body is moving upwards, acceleration a = - g (acceleration due to gravity) Let the body be thrown from point O. Thus, $v_1,v_2$ and $v_3$ are the final velocities at the end of distances OA , OB and OC respectively. The corresponding figure is shown below : Using the third equation of motion : $v^2-u^2$ = 2aS For distance OA: $v_1^2-u^2$ = - 2g (OA) ⇒ OA = $\frac{(u-2)^2-u^2}{-2g}$ = $\frac{3u^2}{8g}$ for distance OB : $v_2^2-u^2$ = - 2g (OB) ⇒ OB = $\frac{\left(\frac{u}{3}\right)^2-u^2}{-2g}$ = $\frac{8u^2}{18g}$ for distance OC : $v_3^2-u^2$ = - 2g (OC) ⇒ OC = $\frac{\left(\frac{u}{4}\right)^2-u^2}{-2g}$ = $\frac{15u^2}{32g}$ Thus, AB = OB - OA ⇒ AB = $\frac{8u^2}{18g} - \frac{3u^2}{8g}$ = $\frac{5u^2}{72g}$ And BC = OC - OB ⇒ BC = $\frac{15u^2}{32g} - \frac{8u^2}{18g}$ = $\frac{7u^2}{288g}$ Therefore, $\frac{AB}{BC}$ = $\frac{\frac{5u^2}{72g}}{\frac{7u^2}{288g}}$ = $\frac{20}{7}$ Hence, option 'D' is correct.