# A baseball is hit off a 1 meter high tee at an angle of 36.9 degrees and at a speed of 27.4 m/s.How far to the nearest tenth of a meter does it travel horizontally?

Apr 14, 2018

$x \approx 74.9$ m/s is the horizontal distance the ball travels.

#### Explanation:

The equation that describes the motion of a projectile near the surface of the earth neglecting air resistance is

$y = {y}_{0} + x \tan \theta - \frac{g {x}^{2}}{2 {v}^{2} {\cos}^{2} \theta}$

where

$y$ = the vertical distance or the projectile from the ground.

${y}_{0}$ = the height from which the projectile was launched = 1 m in this problem.

$x$ = the horizontal position of the projectile.

$\theta$ = the angle that the direction of the projectile's initial velocity makes with the ground = ${36.9}^{\circ}$ in this problem.

$g$ = the gravitational acceleration at the surface of the earth $\approx 9.8 \frac{m}{s} ^ 2$.

$v$ = the initial magnitude of the velocity of the projectile = 27.4 m/s in this problem.

We want to know the $x$ value when $y = 0$. This means we need to set $y = 0$ and then solve for $x$.

$0 = {y}_{0} + x \tan \theta - \frac{g {x}^{2}}{2 {v}^{2} {\cos}^{2} \theta}$

Use the quadratic formula to solve for $x$.

$x = \frac{- \tan \theta \pm \sqrt{{\tan}^{2} \theta - 4 \left(- \frac{g}{2 {v}^{2} {\cos}^{2} \theta}\right) {y}_{0}}}{- 2 \frac{g}{2 {v}^{2} {\cos}^{2} \theta}}$

$= \frac{- \tan \theta \pm \sqrt{{\tan}^{2} \theta + \frac{2 g {y}_{0}}{{v}^{2} {\cos}^{2} \theta}}}{- \frac{g}{{v}^{2} {\cos}^{2} \theta}}$

$= \frac{- \tan \left({36.9}^{\circ}\right) \pm \sqrt{{\tan}^{2} \left({36.9}^{\circ}\right) + \frac{2 \cdot 9.8 \cdot 1}{{\left(27.4\right)}^{2} {\cos}^{2} \left({36.9}^{\circ}\right)}}}{- \frac{9.8}{{\left(27.4\right)}^{2} {\cos}^{2} \left({36.9}^{\circ}\right)}}$

The two values we calculate are $x \approx - 1.3$ and $x \approx 74.9$m/s.

$x \approx 74.9$ m/s is the horizontal distance the ball travels.