1 Answer

BrokenBenchmark · Answer 1 · 2023-10-25T23:09:12+0000

By third equation of motion -

$\green{ \underline { \boxed{ \sf{v^2-u^2=2aS}}}}$

where

Putting Values to find final velocity of mass before hitting the ground-

$\begin{gathered}\\\implies\quad \sf v^2-(0)^2=2* g * 4 \quad (g = acceleration \: due \:to \: gravity) \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf v^2=2* 9.8 * 4\quad (g= 9.8 \:m/s) \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf v=√(78.6) \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf v= 8.86 \:m/s \\\end{gathered}$

Now finding the momentum of the mass at that moment -

$\green{ \underline { \boxed{ \sf{Momentum= mass * velocity}}}}$

$\begin{gathered}\\\implies\quad \sf Momentum= 21.2 * 8.86 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf Momentum= 187.8 \:kgms^(-1) \\\end{gathered}$

$\longrightarrow$ The momentum of the mass just before it hits the ground is 187.8 kgm/s

$\\$

$\therefore \sf Option \: B) \: is \:correct$ ✔️

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