Question

A small sphere is projected with a velocity of $3m{s}^{-1}$ in a direction ${60}^{\circ }$ from the horizontal $y-axis$, on the smooth inclined plane (Fig.). The motion of sphere takes place in the $x-y$ plane.
Calculate the magnitude $v$ of its velocity after $2s$.

Answer 1

The motion of the sphere takes place in a plane; the $x-$ and $y-$ components of its acceleration are $A_x=-g\sin {30}^{\circ },a_y=0$
The $x-$ and $y-$ components of the sphere's velocity at time $t=2s$ are
$V_x=v_{0x}-a_{x}t=3\sin {60}^{\circ }-g\sin {30}^{\circ }\times 2=0.1m{s}^{-1}$
$V_y=v\cos {60}^{\circ }=constant=1.5m{s}^{-1}$
So the magnitude of sphere's velocity is
$|\overrightarrow{v}|=\sqrt{v_x^2+v_y^2}=\sqrt{(0.1{)}^2+(1.5{)}^2}=1.5m{s}^{-1}$.