Question

A car driver is trying to jump across a path as shown in figure by driving horizontally off a cliff ' $X$ ' at the speed 10 $m/s$. When he touches peak $Z$ (ignore air resistance), what would be speed?
$\text{ (use }g=10m/s^2\text{ ) }$

• A. $30m/s$
• B. $40m/s$
• C. $15m/s$
• D. $50m/s$

Answer 1

Answer: (A)

Given, speed of car at $X,v_x=10m/s$.
Acceleration due to gravity, $g=10m{s}^{-2}$
Let, mass of car be $m$.

Speed of car at $X$ in downward direction $\left(v_1\right)$ be $0m{s}^{-1}$,
Speed of car at $Z$ in downward direction be $v_2$ and $v$ be the speed of car at $Z$,
$h_1$ and $h_2$ are the position of $X$ and $Z$ from ground i.e., $100m$ and $60m$, respectively.
Now, by using law of conservation of energy
$\frac{1}{2}m{v}_2^2=mg\left(h_1-h_2\right)$
$\Rightarrow v_2=\sqrt{2g\left(h_1-h_2\right)}$
$=\sqrt{2\times 10(100-60)}$
$=\sqrt{2\times 10\times 40}$
$=\sqrt{800}$
$=20\sqrt{2}m{s}^{-1}$
$v=10\hat{i}+20\sqrt{2}\hat{j}$
$|v|=\sqrt{10^2+(20\sqrt{2}{)}^2}$
$=\sqrt{100+400\times 2}$
$=\sqrt{900}=30m{s}^{-1}$