A batter hits a fly ball which leaves the bat 0.95 m above the ground at an angle of 61° with an initial speed of 30 m/s heading toward centerfield. Ignore air resistance.

(a) How far from home plate would the ball land if not caught?

(b) The ball is caught by the centerfielder, who starts running straight toward home plate at a constant speed from 105 m away, at the instant the ball is hit, and makes the catch at ground level. Find his speed.

For this problem, we're looking for the **horizontal distance** the baseball travels given the **magnitude and ****direction** of the initial velocity and the **height** it was batted from. Then we also want to know **how fast the centerfielder runs** to catch the ball as it lands given his **horizontal distance from the launch point**.

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:

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

Projectile Motion: Positive Launch

Projectile Motion: Positive Launch

Projectile Motion: Positive Launch

Projectile Motion: Positive Launch