Question

A small steel ball is dropped from a height of 1.5m into a glycerin jar. The ball just reaches the bottom of the jar 1.5 sec after it was dropped. The height of the glycerin in the jar, if the retardation is 2.66ms${}^{-2}$, is about

• A. 7.0 m
• B. 7.5m
• C. 5.5m
• D. 3.2m

Answer 1

Answer: (C)

5.5m

It is a free fall till ball reaches the upper surface of the glycerine.
Velocity of ball when it reaches the upper surface of glycerine is given as
$v=\sqrt{2gh}=\sqrt{2\times 9.8\times 1.5}$
or
$v^2=29.4m/s$
using the formula $v_f^2=v_i^2-2ah$($v_i$ is the initial velocity)
or
$0=29.4-2\left(2.66\right)h$
or
$h=5.5m$