Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 30.0 m/s and at an angle of 35.1° above the horizontal. You can ignore air resistance Part A For related problemsolving tips and strategies, you may want to view a Video Tutor Solution of Height and range of a projectile la batted baseball At what two times is the baseball at a height of 11.0 m above the point at which it left the bat? Express your answers in seconds. Enter your answers in ascending order separated by a comma. Templates Symbois ufido redo fadet keyboard shortcuts help ti,t2 = Submit Previous Answers Request Answer * Incorrect: Try Again; 4 attempts remaining Part B Calculate the horizontal component of the baseball's velocity at an earlier time calculated in part A. Express your answer in meters per second. Templates Symbols ufuo redo adet keyboard shortcuts help m/s Submit Request Answer Part Calculate the vertical component of the baseball's velocity at an earlier time calculated in part A. Express your answer in meters per second. Templates Symbols uño redo fedet keyboard shortcuts help Submit Request Answer Part D Calculate the horizontal component of the baseball's velocity at a later time calculated in part A. Express your answer in meters per second. Templates symbols undo redp fadet keyboard shortcuts help

Answer 1

$\begin{array}{l}given:\\initial\;velocity,\;v_0=\;30.0\;m/s\\angle\;of\;projection,\;\theta=\;35.1\;degree\\part:A\\height\;above\;ground,\;y=11\;m\\we\;have\;to\;calculate\;the\;time\;\;at\;which\;baseball\;will\;be\;at\;this\;height.\\so\;u\sin g\;equation:\\y=y_0+v_{0y}t+\frac12a_yt^2\\initially,\;y_0=0\;\;\;\;\;\;\;v_{0y}=v_0\;\sin(\theta)\;\;\;\;\;a_y=-g=-9.81\;m/s^2\\putting\;values\\11=0+(30\times\sin(35.1)\times t)\;-\frac12\times9.81\times t^2\\11=17.25015756\;t-4.905\;t^2\\4.905\;t^2-17.25015756\;t+11=0\\\end{array}$

$\begin{array}{l}4.905\;t^2-17.25015756\;t+11=0\\this\;is\;a\;quadratic\;equation\;in\;'t'.\\solution\;of\;this\;type\;of\;equations\;at^2+bt+c=0\;is\\t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\\\Rightarrow t=\frac{-(-17.25015756)\pm\sqrt{{(-17.25015756)}^2-4\times4.905\times11}}{2\times4.905}\\t=\frac{17.25015756\pm9.041454529}{9.81}\\so\;here\;two\;value\;of\;'t'\;are\;possible.\;one\;for\;'+'\;sign\;and\;other\;for\\'-'\;sign.\\t_1=0.8367689\;s\;\;\;\;\;\;\;t_2=2.680083\;s\;\;\;\;\;\;\;(answers)\end{array}$

$\begin{array}{l}part:b\\at\;time\;\;\;\;\;\;t_1=0.8367689\;s\\horizontal\;component\;of\;velocity\;will\;remain\;cons\tan t\;because\;there\;is\;no\;\\acceleration\;along\;horizontal\;axis.\;so\\v_{0x}=v_0\cos(35.1)=30\times\cos(35.1)\\v_{0x}=24.544491\;m/s\;\;\;\;(answer)\end{array}$

$\begin{array}{l}part:c\\component\;of\;vertical\;velocity\;at\;t=t_1\\v_y=v_{0y}+at\\v_y=(v_0\;\sin(35.1)\;-\;gt_1\\v_y=17.25015756-(9.81\times0.8367689)\\v_y=9.041454651\;m/s\end{array}$

$\begin{array}{l}part:d\\at\;time\;\;\;\;t=\;\;t_2\\horizontal\;component\;of\;velocity\;will\;remain\;cons\tan t\;because\;there\;is\;no\;\\acceleration\;along\;horizontal\;axis.\;so\\v_{0x}=v_0\cos(35.1)=30\times\cos(35.1)\\v_{0x}=24.544491\;m/s\;\;\;\;(answer)\end{array}$

$\begin{array}{l}part:e\\when\;it\;moves\;up,\;it\;will\;get\;deaccelerated\;and\;slows\;down\;till\;maximum\;height\;point\\then\;while\;coming\;down\;it\;accelerates.\;so\;at\;time\;t=t_2\;,\;the\;ball\;is\;at\;same\;height\;above\\ground,\;as\;it\;was\;as\;t=t_1.\;so\;it\;will\;have\;the\;same\;velocity\;as\;it\;have\;at\;t=t_1\\v_y=9.041454651\;m/s\end{array}$

$\begin{array}{l}part:f\\when\;baseball\;returns\;to\;the\;level\;where\;it\;left\;the\;bat,\;it\;will\;have\;same\\magnitude\;of\;velocity\;,\;as\;it\;had\;initially.\;\\so\;v=30\;m/s\end{array}$

$\begin{array}{l}part:g\\direction\;of\;baseball\;velocity\;when\;it\;returns\;to\;the\;initial\;level\\the\;direction\;of\;v_y\;will\;be\;downwards.\;so\;it\;will\;be\;taken\;negative\\\tan(\theta)=\frac{v_{0y}}{v_{0x}}=\frac{-17.25015756}{24.544491}=-0.7028118\\\theta=\tan^{-1}\left(-0.7028118\right)\\\theta=-35.1000011\;degree\\negative\;sign\;shows\;angle\;is\;below\;the\;horizontal\;.\;so\;the\;answer\;is\\\theta=35.1000011\;degree\;\;\;\\note:\;dont\;input\;negative\;sign\;.\;its\;just\;showing\;the\;angle\;is\;down\;the\\plane.\;\end{array}$