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?
$\text { (use } g=10 \,m / s ^2 \text { ) }$

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

Answer 1

Answer: (A)

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

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