Given:
the speed of the ball is
[tex]\begin{gathered} v=57\text{ mi/h} \\ v=57\times0.4470\text{ m/s} \\ v=25.48\text{ m/s} \end{gathered}[/tex]
angle at which the ball is projected
[tex]\theta=40^{\circ}[/tex]
Required: the velocity in the vector form
Explanation:
look at the diagram below
velocity in the vector form can be written as
[tex]V=v\cos40^{\circ}i+v\sin40^{\circ}j[/tex]
here, i and j are unit vectors.
plugging all the value in the above relation, and solve
[tex]\begin{gathered} V=25.48\text{ m/s}\times0.7660\text{ }i+25.48\text{ m/s}\times0.6428j \\ V=19.5176i+16.3785j \end{gathered}[/tex]
final answer is
[tex]\begin{gathered} V=19.5176\imaginaryI+16.3785j\text{ m/s} \\ \lbrack V\rbrack=25.4792\text{ m/s} \end{gathered}[/tex]
