🤓 Based on our data, we think this question is relevant for Professor Efthimiou's class at UCF.

For this problem, we're looking for the **launch height** and **final speed** of the golf ball given the **magnitude and ****direction** of the initial velocity and the **horizontal distance** it has to travel.

For **projectile motion problems in general**, we'll follow these steps to solve:

- Identify the
and__target variable__for each direction—remember that__known variables__*only*(Δ**3**of the**5**variables*x*or Δ*y*,*v*_{0},*v*,_{f}*a*, and*t*)*are needed*for each direction. Also, it always helps to sketch out the problem and label all your known information! __Choose a UAM__—sometimes you'll be able to go directly for the target variable, sometimes another step will be needed in between.**equation**for the target (or intermediate) variable, then**Solve**the equation__substitute known values__and__calculate__the answer.

The four UAM (kinematics) equations are:

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

In our coordinate system, the **+ y-axis is pointing upwards** and the

For projectiles with a** positive launch angle**, we __also__ need to know how to decompose a velocity vector into its *x*- and *y*-components:

$\overline{)\begin{array}{rcl}{\mathit{v}}_{\mathbf{0}\mathit{x}}& {\mathbf{=}}& \mathbf{\left|}{\stackrel{\mathbf{\rightharpoonup}}{\mathit{v}}}_{\mathbf{0}}\mathbf{\right|}\mathbf{}\mathbf{cos}\mathbf{}\mathit{\theta}\\ {\mathit{v}}_{\mathbf{0}\mathit{y}}& {\mathbf{=}}& \mathbf{\left|}{\stackrel{\mathbf{\rightharpoonup}}{\mathit{v}}}_{\mathbf{0}}\mathbf{\right|}\mathbf{}\mathbf{sin}\mathbf{}\mathit{\theta}\end{array}}$

And the equations to find the magnitude and direction of a velocity from the components:

$\overline{)\mathbf{\left|}\stackrel{\mathbf{\rightharpoonup}}{\mathit{v}}\mathbf{\right|}{\mathbf{=}}\sqrt{{{\mathit{v}}_{\mathit{x}}}^{\mathbf{2}}\mathbf{+}{{\mathit{v}}_{\mathit{y}}}^{\mathbf{2}}}}$

$\overline{){\mathbf{tan}}{\mathbf{}}{\mathit{\theta}}{\mathbf{=}}\frac{{\mathit{v}}_{\mathit{y}}}{{\mathit{v}}_{\mathit{x}}}}$

A golfer tees off from the top of a rise, giving the golf ball an initial velocity of 43.0 m/s at an angle of 30.0° above the horizontal. The ball strikes the fairway a horizontal distance of 180 m from the tee. Assume the fairway is level. (a) How high is the rise above the fairway? (b) What is the speed of the ball as it strikes the fairway?