Kinematic equations:

$\overline{)\mathbf{}{{\mathit{v}}}_{{\mathit{f}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{{\mathit{v}}}_{{\mathbf{0}}}{\mathbf{}}{\mathbf{-}}{\mathit{g}}{\mathit{t}}\phantom{\rule{0ex}{0ex}}{\mathbf{\u2206}}{\mathit{y}}{\mathbf{=}}{\mathbf{}}\mathbf{\left(}\frac{{\mathit{v}}_{\mathit{f}}\mathbf{+}{\mathit{v}}_{\mathbf{0}}}{\mathbf{2}}\mathbf{\right)}{\mathit{t}}\phantom{\rule{0ex}{0ex}}{\mathbf{\u2206}}{\mathit{y}}{\mathbf{=}}{\mathbf{}}{{\mathit{v}}}_{{\mathbf{0}}}{\mathit{t}}{\mathbf{-}}\frac{\mathbf{1}}{\mathbf{2}}{\mathit{g}}{{\mathit{t}}}^{{\mathbf{2}}}\phantom{\rule{0ex}{0ex}}{\mathbf{}}{{{\mathit{v}}}_{{\mathit{f}}}}^{{\mathbf{2}}}{\mathbf{=}}{\mathbf{}}{{{\mathit{v}}}_{{\mathbf{0}}}}^{{\mathbf{2}}}{\mathbf{}}{\mathbf{-}}{\mathbf{2}}{\mathit{g}}{\mathbf{\u2206}}{\mathit{y}}}$

**Part A.**

From the fourth kinematic equation:

$\begin{array}{rcl}{{{\mathbf{v}}}_{{\mathbf{fy}}}}^{\mathbf{2}}& \mathbf{=}& {{{\mathbf{v}}}_{{\mathbf{0}}{\mathit{y}}}}^{\mathbf{2}}\mathbf{}\mathbf{-}\mathbf{2}\mathbf{g}\mathbf{\u2206}\mathbf{y}\\ \mathbf{\u2206}\mathbf{y}& \mathbf{=}& \frac{{{{\mathbf{v}}}_{{\mathbf{0}}{\mathit{y}}}}^{\mathbf{2}}\mathbf{-}{{{\mathbf{v}}}_{{\mathbf{fy}}}}^{\mathbf{2}}}{\mathbf{2}\mathbf{g}}\end{array}$

A projectile is fired with speed v_{0} at an angle θ from the horizontal as shown in the figure below.

Part A. Find the highest point in the trajectory, H. Express the highest point in terms of the magnitude of the acceleration due to gravity g, the initial velocity v_{0}, and the angle θ.

Part B. What is the range of the projectile, R? Express the range in terms of v_{0}, θ, and g.

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