2D vectors Components:

$\overline{)\begin{array}{rcl}{\mathit{A}}_{\mathit{x}}& {\mathbf{=}}& \mathbf{\left|}\stackrel{\mathbf{\rightharpoonup}}{\mathit{A}}\mathbf{\right|}\mathbf{}\mathbf{cos}\mathbf{}\mathit{\theta}\\ {\mathit{A}}_{\mathit{y}}& {\mathbf{=}}& \mathbf{\left|}\stackrel{\mathbf{\rightharpoonup}}{\mathit{A}}\mathbf{\right|}\mathbf{}\mathbf{sin}\mathbf{}\mathit{\theta}\end{array}}$

The maximum height reached:

$\overline{){{\mathbf{h}}}_{\mathbf{m}\mathbf{a}\mathbf{x}}{\mathbf{=}}\frac{{{\mathbf{v}}_{\mathbf{0}\mathit{y}}}^{\mathbf{2}}}{\mathbf{2}\mathbf{g}}}$

**(a)**

From the vectors components equation:

v_{0}_{x} = v_{0}cosθ

The magnitude of the ball's initial horizontal velocity is v_{0}cosθ.

During a baseball game, a baseball is struck at ground level by a batter. The ball leaves the baseball bat with an initial velocity v_{0} = 25 m/s at an angle θ = 15° above horizontal. Let the origin of the Cartesian coordinate system be the ball's position the instant it leaves the bat. Air resistance may be ignored throughout this problem.

**(a)** Express the magnitude of the ball's initial horizontal velocity v_{0x} in terms of v_{0} and θ.**(b)** Express the magnitude of the ball's initial vertical velocity v_{0y} in terms of v_{0} and θ.**(c)** Find the ball's maximum vertical height h_{max} in meters above the ground.

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