Question

A boy of height $1.5\text{m}$ is moving on a skateboard due east with velocity $4\text{m/s}$. He throws a coin vertically upward with a velocity of $3\text{m/s}$ relative to himself. Find the total displacement of the coin relative to ground till it comes to the hand of the boy. Take $g=10{\text{m/s}}^2$.

• A. $0\text{m}$
• B. $\sqrt{{0.45}^2+{12}^2}\text{m}$
• C. $\frac{12}{5}\text{m}$
• D. $\sqrt{{0.45}^2+{2.4}^2}\text{m}$

Answer 1

The correct option is C $\frac{12}{5}\text{m}$

Path of the coin will be a parabola due to its two components of velocity.
From relative velocity equation,
${\overrightarrow{u}}_{cg}={\overrightarrow{u}}_{cb}+{\overrightarrow{u}}_{bg}$ where
${\overrightarrow{u}}_{cg}=$ Initial velocity of coin w.r.t ground
${\overrightarrow{u}}_{cb}=$ Initial velocity of coin w.r.t boy
${\overrightarrow{u}}_{bg}=$ Initial velocity of boy w.r.t ground

$\Rightarrow$ Initial velocity of coin w.r.t ground
${\overrightarrow{u}}_i=u_x\hat{i}+u_y\hat{j}=4\hat{i}+3\hat{j}\text{m/s}$

Displacement of coin will be along $x$ direction only, when it reaches back to the boy's hand at time $t$
Time of flight $t=\frac{2u_y}{g}=\frac{6}{10}=\frac{3}{5}\text{s}$
and Displacement along x-direction
$X=u_{x}t$
$\therefore X=4\times \frac{3}{5}=\frac{12}{5}\text{m}$