Question

A particle leaves the origin with an initial velocity $\overset{ \rightarrow }{\text{v}} = \left(\text{3.00} \hat{\text{i}}\right) \left(\text{ms}\right)^{- 1}$ and a constant acceleration $\overset{ \rightarrow }{\text{a}} = \left(- \text{1.00} \hat{\text{i}} - \text{0.5} \hat{\text{j}}\right) \left(\text{ms}\right)^{- 2}$ . When the particle reaches it maximum $x$ -coordinate, what is the magnitude of its velocity (in m/s) in $y$ -direction?

Answer 1

Velocity along $x$ direction is
$v_{x} = v_{0 2} + \textit{at}$
$v_{x} = 3 + \left(- 1\right) t$
so when particle reach maximum $x$ coordinate $v_{x} = 0 \Rightarrow t = 3 sec$
similarly $v_{y} = v_{0 y} + a_{y} t$
$v_{y} = 0 + \left(- \frac{1}{2}\right) t$
$\left(v_{y}\right)_{at\, t = 3}=-\frac{3}{2}=-1.5 \, m \, s^{- 1}$