Uniform accelerated motion (UAM) equations:

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

**(1)**

Range = 10.0 m

- H
_{max}= 3.0 m - v
_{f,y}= 0 m/s - g = 9.8 m/s
^{2} - v
_{0,y}= ? - t = ?

We'll use the motion in the y-direction to determine v_{0,y} and flight time.

Using the last UAM equation:

A running mountain lion can make a leap 10.0 m long, reaching a maximum height of 3.0 m.

1. What is the speed of the mountain lion just as it leaves the ground?

2. At what angle does it leave the ground?

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