The range, R is expressed as:

$\overline{){\mathbf{R}}{\mathbf{=}}\frac{{{\mathbf{v}}_{\mathbf{0}}}^{\mathbf{2}}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\left(}\mathbf{2}\mathbf{\theta}\mathbf{\right)}}{\mathbf{g}}}$

Time of flight of a projectile projected with some angle above the ground:

$\overline{){\mathbf{T}}{\mathbf{=}}\frac{\mathbf{2}{\mathbf{v}}_{\mathbf{0}}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta}}{\mathbf{g}}}$

We'll use the kinematic equation:

$\overline{){\mathbf{\u2206}}{\mathit{y}}{\mathbf{=}}{\mathbf{}}{{\mathit{v}}}_{{\mathbf{0}}}{\mathit{t}}{\mathbf{+}}{\frac{1}{2}}{\mathit{g}}{{\mathit{t}}}^{{\mathbf{2}}}}$

**Part A**

We need to use the equation for the time of flight, but we don't have v_{0}. We can find v_{0} from the expression for range as:

$\begin{array}{rcl}{\mathbf{v}}_{\mathbf{0}}& \mathbf{=}& \sqrt{\frac{\mathbf{R}\mathbf{g}}{\mathbf{sin}\mathbf{\left(}\mathbf{2}\mathbf{\theta}\mathbf{\right)}}}\\ & \mathbf{=}& \sqrt{\frac{\mathbf{\left(}\mathbf{220}\mathbf{\right)}\mathbf{(}\mathbf{9}\mathbf{.}\mathbf{8}\mathbf{)}}{\mathbf{sin}\mathbf{\left[}\mathbf{\right(}\mathbf{2}\mathbf{\left)}\mathbf{\right(}\mathbf{45}\mathbf{\xb0}\mathbf{\left)}\mathbf{\right]}}}\end{array}$

v_{0} = 46.43 m/s

An arrow is shot at an angle of θ = 45° above the horizontal. The arrow hits a tree a horizontal distance D = 220m away, at the same height above the ground as it was shot. The height from which the arrow is shot is 6 m. Use g=9.8m/s2 for the magnitude of the acceleration due to gravity.

**Part A**

Find *t*_{a}, the time that the arrow spends in the air.

**Part B**

How long after the arrow was shot should the apple be dropped, in order for the arrow to pierce the apple as the arrow hits the tree?

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