Question

A batter hits a baseball so that it leaves the bat with an initial velocity of 28.0m/s at an angle of 35°. What is the maximum height, the time it is in the air, and the range?

Single line text.

Answer 1

The information given is

• The initial velocity is $V_{o}=28.0\;\text{m}/\text{s}$
• The launch angle is $\theta=35^{0}$
• Gravity acceleration is $a_{y}=-9.81\;\text{m}/\text{s}^{2}$
• The starting position for the move is $x_{o}=0\;\text{m}\;\text{and}\;y_{o}=0\;\text{m}$

\\

Components of the initial velocity.

Horizontal component.

$cos(35^{0})=\dfrac{V_{ox}}{V_{o}}\\ cos(35^{0})\;V_{o}=V_{0x}\\ V_{0x}=V_{o}\;cos(35^{0})\\ V_{ox}=28.0\;\text{m}/\text{s}\times cos(35^{0})\\ V_{ox}=22.9\;\text{m}/\text{s}$

Vertical component.

$sin(35^{0})=\dfrac{V_{oy}}{V_{o}}\\ sin(35^{0})\;V_{o}=V_{0y}\\ V_{0y}=V_{o}\;sin(35^{0})\\ V_{oy}=28.0\;\text{m}/\text{s}\times sin(35^{0})\\ V_{oy}=16.1\;\text{m}/\text{s}$

\\

The maximum height is calculated with the following equation.

$\Delta Y_{max}=\dfrac{ V_{fy}^{2}-V_{oy}^{2}}{2\;a_{y}}$

\\

Where.

$V_{oy}$ is the initial vertical velocity.

$V_{fy}$ is the velocity at the maximum height.

$a_{y}$ is the acceleration of gravity.

\\

Remember that at the top, the vertical component of velocity is zero.

$V_{fy}=0\;\text{m}/\text{s}$

\\

Evaluating numerically.

$\Delta Y_{max}=\dfrac{ V_{fy}^{2}-V_{oy}^{2}}{2\;a_{y}}\\ \Delta Y_{max}=\dfrac{ (0\;\text{m}/\text{s} )^{2}-(16.1\;\text{m}/\text{s})^{2}}{2\times-9.81\;\text{m}/\text{s}^{2}}\\ \Delta Y_{max}=13.2\;\text{m}$

\\

!!Answer.

The maximum height is $\displaystyle \color{red}{\boxed{\Delta Y_{max}=13.2\;\text{m}}}$

\\

Flight time is calculated using the following equation.

$V_{fy}=V_{oy}+a_{y}\;t$

\\

where

$V_{oy}$ is the initial vertical velocity.

$V_{fy}$ is the final vertical velocity.

$a_{y}$ is the vertical acceleration.

\\

Remember that if the movement is symmetrical, the vertical velocity with which it reaches the ground is of the same magnitude but in the opposite direction

$V_{fy}=-V_{oy}$ .

\\

Obtaining the expression for time.

$V_{fy}=V_{oy}+a_{y}\;t\\ V_{oy}+a_{y}\;t=V_{fy}\\ a_{y}\;t=V_{fy}-V_{oy}\\ t=\dfrac{V_{fy}-V_{oy}}{a_{y} }\\ \text{if}\;V_{fy}=-V_{oy}\\ t=\dfrac{-V_{oy}-V_{oy}}{a_{y}}\\ t=\dfrac{-2\;V_{oy}}{a_{y}}$

\\

Evaluating numerically.

$t=\dfrac{-2\;V_{oy}}{a_{y}}\\ t=\dfrac{-2\times 16.1\;\text{m}/\text{s}}{-9.81\;\text{m}/\text{s}^{2}}\\ t=3.28\;\text{s}$

\\

!!Answer.

Flight time is $\displaystyle \color{red}{\boxed{t=3.28\;\text{s}}}$

\\

The distance traveled is given by the following equation.

$X=X_{0}+V_{ox}\;t$

\\

where.

$V_{ox}$ is the horizontal velocity.

$t$ Is time.

$X_{o}$ is the starting position.

\\

Evaluating numerically

$X=X_{0}+V_{ox}\;t\\ X=0\;\text{m}+22.9\;\text{m}/\text{s}\times 3.28\;\text{s}\\ x=75.1\;\text{m}$

\\

!!Answer

The maximum horizontal reach is $\displaystyle \color{red}{\boxed{x=75.1\;\text{m}}}$