# If a baseball player throws a ball at 35.0 m/s what is its maximum range?

Feb 7, 2017

$124.87 m$, rounded to two decimal places.

#### Explanation:

The maximum horizontal distance traveled by the base ball is the horizontal distance it covers when it returns to its initial height. Let it be $\left(y = 0\right)$ as shown below.

If the ball is thrown making an angle $\theta$ with the horizontal with velocity ${v}_{0}$
Using the kinematic equation
$s = u t + \frac{1}{2} g {t}^{2}$,

Noting that acceleration due to gravity $g = 9.81 m {s}^{-} 2$ is opposing the direction of motion. we get for vertical component of velocity
$0 = {v}_{0} t \setminus \sin \theta - \setminus \frac{1}{2} g {t}^{2}$
$\implies 0 = {v}_{0} \setminus \sin \theta - \setminus \frac{1}{2} g t$

Time $t$ taken to return to ground or to travel distance $d$ is found as
t = (2 v_0 sin ⁡ θ) / g ....(1)

Ignoring air resistance,
Distance $d = \text{Horizontal component of velocity"xx"time}$
$= \left({v}_{0} \cos \setminus \theta\right) t$ .....(2)

Inserting value of $t$ from (1) in (2) we get
d=(v_{0}cos\theta )(2 v_0 sin ⁡ θ) / g
$\implies d = {v}_{0}^{2} / g 2 \sin \theta \cos \setminus \theta$
$\implies d = {v}_{0}^{2} / g \sin 2 \theta$
For its maximum range $\sin 2 \theta = 1$
$\implies {d}_{\max} = {v}_{0}^{2} / g$

Taking Inserting values we get
${d}_{\max} = {\left(35.0\right)}^{2} / 9.81$
$\implies {d}_{\max} = 124.87 m$, rounded to two decimal places.