Question

A boy of height 1.5 m, making move on a skateboard due east with velocity 4 m s${}^{-1}$, throws a coin vertically up with a velocity of 3 m s${}^{-1}$ relative to himself.
a. Find the total displacement of the coin relative to ground till it comes to the hand of the boy.
b. What is the maximum height attained by the coin w.r.t to ground?

Answer 1

The path of the coin as seen from boy A is a straight line (up and down), but that seen from ground observer B is parabola.

Velocity of coin, ${\overrightarrow{v}}_{coin}={\overrightarrow{v}}_{coin,boy}+{\overrightarrow{v}}_{boy}{\overrightarrow{v}}_{coin}=4\hat{i}+3\hat{k}\left(m{s}^{-1}\right)$

Acceleration of coin , ${\overrightarrow{a}}_{coin}=-10\hat{k}\left(m{s}^{-2}\right)$

The displacement of coin in $z$-direction will be zero.

Using $\overrightarrow{s}=\overrightarrow{u}t+\frac{1}{2}\overrightarrow{a}{t}^2$ in vertical direction ($z$-direction),

$0=3\left(T\right)-\frac{1}{2}\times 10\times (T{)}^2$

($T$ = Time of flight of coin)

or $0=T(3-5T)\Rightarrow T=\frac{3}{5}s$

The displacement of the coin will be only $x$-direction. Let the displacement in $x$-direction be $x$. Then

$x=v_x\times T=4\times \frac{3}{5}=\frac{12}{5}m$

At maximum height, the final component of coin velocity in vertical direction will be zero.

Using $v^2=u^2+2as$ in vertical direction,

$0=(3{)}^2-2.10H\Rightarrow H=\frac{9}{20}m$

Hence, maximum height reached by coin

$H_{max}$ = Height of boy +$H=1.5+\frac{9}{20}=\frac{39}{20}m$

a. Hence the displacement of coin will be in horizontal direction and will be equal to $12/5$ m.

b. The maximum height attained by coin will be $39/20$ m.