91Ó°ÊÓ

The vector form of the velocity is expressed as follows:

$\begin{array}{l}\stackrel{→}{v}=v_x\widehat{i}+v_y\widehat{j}\\ =v\cos \mathrm{θ}\widehat{i}+v\sin \mathrm{θ}\widehat{j}\end{array}$

Here, $v_x$is the horizontal component of the velocity, $v_y$is the vertical component of the velocity, anddata-custom-editor="chemistry" $\mathrm{θ}$ is the angle.

The horizontal component of the velocity is expressed as follows:

$v_x=v\cos \mathrm{θ}$

Substitute $2.5\text{m}/\text{s}$for $v_x$and $3\text{m}/\text{s}$for $v$and solve for $\mathrm{θ}$.

$\begin{array}{l}2.5\text{m}/\text{s}=(3\text{m}/\text{s})\cos \mathrm{θ}\cos \mathrm{θ}\\ =\frac{2.5\text{m}/\text{s}}{3\text{m}/\text{s}}\mathrm{θ}\\ ={\cos }^{−1}\left(\frac{2.5\text{m}/\text{s}}{3\text{m}/\text{s}}\right)\\ ={33.55}^{°}\end{array}$

Therefore, the angle after rounding off to two significant digits is${34}^{°}$.

The vertical component of the velocity is expressed as follows:

$v_y=v\sin \mathrm{θ}$

Substitute $3\text{m}/\text{s}$$v$for and ${33.55}^{°}$for$\mathrm{θ}$.

$\begin{array}{l}{v}_y=v\sin \mathrm{θ}\\ =(3\text{m}/\text{s})\sin {33.55}^{°}\\ =1.658\text{m}/\text{s}\end{array}$

Therefore, the vertical component of the velocity after rounding off to two significant digits is$1.7\text{m}/\text{s}$.