Consider the position equation:

$\overline{){\mathbf{y}}{\mathbf{=}}{{\mathbf{y}}}_{{\mathbf{0}}}{\mathbf{+}}{\mathbf{}}{{\mathbf{v}}}_{{\mathbf{0}}}{\mathbf{t}}{\mathbf{+}}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{g}}{{\mathbf{t}}}^{{\mathbf{2}}}}$

**(a)**

The initial velocity, v_{0} = 0

So we have,

$\begin{array}{rcl}\mathbf{y}& \mathbf{=}& \mathbf{0}\mathbf{+}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{g}{\mathbf{t}}^{\mathbf{2}}\\ \mathbf{t}& \mathbf{=}& \sqrt{\frac{\mathbf{2}\mathbf{y}}{\mathbf{g}}}\\ & \mathbf{=}& \sqrt{\frac{\mathbf{\left(}\mathbf{2}\mathbf{\right)}\mathbf{(}\mathbf{-}\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{)}}{\mathbf{-}\mathbf{9}\mathbf{.}\mathbf{8}}}\end{array}$

A ball is thrown horizontally from the top of a 60.0-m building and lands 100.0 m from the base of the building. Ignore air resistance.

(a) How long is the ball in the air?

(b) What must have been the initial horizontal component of the velocity?

(c) What is the vertical component of the velocity just before the ball hits the ground?

(d) What is the velocity (including both the horizontal and vertical components) of the ball just before it hits the ground?

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