A small steel ball is dropped from a height of $1.5 \mathrm{~m}$ into a glycerine jar. The ball reaches the…
A small steel ball is dropped from a height of $1.5 \mathrm{~m}$ into a glycerine jar. The ball reaches the bottom of the jar 1.5 second after it was dropped. If the retardation is $2.66 \mathrm{~m} / \mathrm{s}^2$, the height of the glycerine in the jar is about (acceleration due to gravity $g=9.8 \mathrm{~m} / \mathrm{s}^2$ )
$7.0 \mathrm{~m}$
$7.5 \mathrm{~m}$
$5.5 \mathrm{~m}$
$3.2 \mathrm{~m}$
Solution
The velocity of the ball after it has been dropped from height till it reaches the glycerine surface is
$\mathrm{v}_{\mathrm{i}}^2=0+2 \mathrm{gh} \quad \quad \ldots .\left(\because \mathrm{v}^2-\mathrm{u}^2=2 \mathrm{gh}\right)$
$v_i^2=2 \times 9.8 \times 1.5$
$v_i^2=29.4$
The velocity of the ball after it enters glycerine is $v_f^2=v_i^2-2 g h$
$0=29.4-(2 \times 2.66 \times h)$
$\therefore \quad \mathrm{h}=\frac{29.4}{2 \times 2.66}=5.5 \mathrm{~m}$