# A Baseball is hit when it is 2.7 feet above the ground. It leaves the bat with n initial speed of...

## Question:

A Baseball is hit when it is 2.7 feet above the ground. It leaves the bat with n initial speed of 150 ft/sec making an angle of {eq}18^{\circ} {/eq} with the horizontal. Assume a drag coefficient k=6.12. find the range and the flight time of the ball

For projection with linear drag

{eq}x=x_0+ \frac{v_0}{k}(1- e^{-kt}) cos \alpha \\ y= y_0+ \frac{v_0}{k}(1- e^{-kt}) sin \alpha +\frac{8}{k}(1-kt-e^{-kt}) {/eq}

Where k is the drag coefficient and v_0 and \alpha are the projectiles initial speed and the lauch angle, and g is the acceleration of the gravity (32ft/sec^2)

a. Flight time=2.93 sec; range=351.6 feet

b. Flight time=2.81 sec; range=340.3 feet

c. Flight time=3.16 sec; range=362.5 feet

d. Flight time=2.01 sec; range=331.6 feet

## Equations

When a polynomial is equated with a zero, it is said to be an equation. Similarly, when a function is placed in front of zero with an 'equal to' sign in between them, it is also called as an equation.

## Answer and Explanation:

Given,

{eq}x_0 = 0 \: ft\\ y_0 = 2.7 \: ft\\ \alpha = 18^{\circ}\\ v_0 = 150 \: ft/sec\\ g = 32 \: ft/sec^2\\ k = 6.12 {/eq}

To find the flight time:

Consider the given equation of vertical co-ordinate;

{eq}y= y_0+ \frac{v_0}{k}(1- e^{-kt}) sin \alpha +\frac{8}{k}(1-kt-e^{-kt}) {/eq}

Substituting the given values, we get;

{eq}y= 2.7 + \frac{150}{6.12}(1- e^{-6.12t}) sin 18 +\frac{8}{6.12}(1-6.12t-e^{-6.12t})\\ y =2.7 + 7.574(1- e^{-6.12t}) +1.307(1-6.12t-e^{-6.12t}) \\ {/eq}

Equating *y* to *0*, we get;

{eq}\begin{align*} y &= 0\\ 2.7 + 7.574(1- e^{-6.12t}) +1.307(1-6.12t-e^{-6.12t}) &= 0\\ 2.7 + 7.574 - 7.574e^{-6.12t} + 1.307 - 8t - 1.307e^{-6.12t} &= 0\\ 11.581 - 8.881e^{-6.12t} -8t &= 0\\ 0.9t - 1.304-e^{-6.12t} =0 \end{align*} {/eq}

on solving ,we get

{eq}\begin{align*} t = 1.12\;{\rm{sec}} \end{align*} {/eq}

Put this value into equation of range

{eq}\begin{align*} x &= {x_0} + \dfrac{{{x_0}}}{k}\left( {1 - {e^{ - kt}}} \right)\cos \alpha \\ x - {x_0} &= \dfrac{{{v_0}}}{k}\left( {1 - {e^{ - kt}}} \right)\cos \alpha \\ x &= \dfrac{{150}}{{6.12}}\left( {1 - {e^{ 6.12\left( {1.12} \right)}}} \right)\cos {18^ \circ }\\ x &= 7.574 \times \left( {1 - {e^{ 6.12\left( {1.12} \right)}}} \right) \times \cos {18^ \circ }\\ x &= 2860.89\;{\rm{ft}} \end{align*} {/eq}

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from Explorations in Core Math - Algebra 1: Online Textbook Help

Chapter 9 / Lesson 7