# Need help with a basic table

I am a beginner. I have been attempting to make a table for the last two days but I am not getting it. I need help with this table.

Some of my attempts

\begin{tabular}{|c|c|c|c|c|}
\hline
$t$ & $y$ & $y$ \text{in terms of} $y_0[=(-\frac 12) g\tau^2]$ &\text{Distance traversed in successive intervals}& \text{Ratio of distances traversed} \
\hline
$0$ & $0$ & $0$ & $:$ & $:$  \
$\tau$  & $-(\frac 12) g \tau^2$ & $y_0$ & $y_0$ & $1$\
$2 \tau$ & $-4(\frac 12) g \tau^2$ & $4y_0$ & $3y_0$ & $3$ \
$3 \tau$ & $-9(\frac 12) g \tau^2$ & $9y_0$ & $5y_0$ & $5$  \
$4\tau$ & $-16(\frac 12) g \tau^2$ & $16y_0$ &$7y_0$ & $7$  \
$5 \tau$ & $-25(\frac 12) g \tau^2$ & $25y_0$ & $9y_0$ & $9$ \
$6 \tau$ & $-36(\frac 12) g \tau^2$& $36y_0$ & $11y_0$ & $11$
\end{tabular}


.

\begin{tabular}{|c|c|c|c|c|}
\hline
$t$ & $y$ & $y$ \text{in terms of} $y_0[=\left(-\frac 12\right) g\tau^2]$ &\text{Distance traversed in successive intervals}& \text{Ratio of distances traversed} \
\hline
$0$ & $0$ & $0$ & $:$ & $:$  \
$\tau$  & $-\left(\frac 12\right) g \tau^2$ & $y_0$ & $y_0$ & $1$\
$2 \tau$ & $-4\left(\frac 12\right) g \tau^2$ & $4y_0$ & $3y_0$ & $3$ \
$3 \tau$ & $-9\left(\frac 12\right) g \tau^2$ & $9y_0$ & $5y_0$ & $5$  \
$4\tau$ & $-16\left(\frac 12\right) g \tau^2$ & $16y_0$ &$7y_0$ & $7$  \
$5 \tau$ & $-25\left(\frac 12\right) g \tau^2$ & $25y_0$ & $9y_0$ & $9$ \
$6 \tau$ & $-36\left(\frac 12\right) g \tau^2$& $36y_0$ & $11y_0$ & $11$ \
\hline
\end{tabular}


.

\begin{tabular}{|c|c|c|c|c|}
\hline
$t$ & $y$ & $y$ \text{in terms of} $y_0[=\left(-\frac 12\right) g\tau^2]$ &\text{Distance traversed in successive intervals}& \text{Ratio of distances traversed} \\
\hline
$0$ & $0$ & $0$ & $\:$ & $\:$  \\
$\tau$  & $-\left(\frac 12\right) g \tau^2$ & $y_0$ & $y_0$ & $1$\\
$2 \tau$ & $-4\left(\frac 12\right) g \tau^2$ & $4y_0$ & $3y_0$ & $3$ \\
$3 \tau$ & $-9\left(\frac 12\right) g \tau^2$ & $9y_0$ & $5y_0$ & $5$  \\
$4\tau$ & $-16\left(\frac 12\right) g \tau^2$ & $16y_0$ &$7y_0$ & $7$  \\
$5 \tau$ & $-25\left(\frac 12\right) g \tau^2$ & $25y_0$ & $9y_0$ & $9$ \\
$6 \tau$ & $-36\left(\frac 12\right) g \tau^2$& $36y_0$ & $11y_0$ & $11$ \\
\hline
\end{tabular}


## 3 Answers

Here is one way to do it:

\documentclass[12pt]{article}
%\usepackage{amsmath}   % in case you need it
\begin{document}
\begin{tabular}{|l|l|p{2cm}|p{2.5cm}|p{3cm}|} % p: col-width
\hline
$t$ &
$y$ &
{$y$ in terms of $y_0=-\frac{1}{2} g \tau^2$} &
{Distance traversed in successive intervals} & % {} for readability
{Ratio of distances traversed}\\% \\ ends the row

\hline  % just two lines as an example
$0$ &$0$ &$0$ &$0$ &$0$\\    % digits in math-font
$6 \tau$ &
$-3 (\frac{1}{2})g \tau^2$ & % watch braces ;-)
{$36 y_0$} & {$11 y_0$} &
{$11$}\\
\hline
\end{tabular}
\end{document}


See https://en.wikibooks.org/wiki/LaTeX/Tables#The_tabular_environment for more details on the tabular environment.

Result:

Unless you know it already, this one is worth spending some hours reading and scanning: https://en.wikibooks.org/wiki/LaTeX .

Find more alternatives for table-design here: https://ctan.org/topic/table?lang=en . May be you enjoy nicematrix https://ctan.org/pkg/nicematrix ?

Have a good journey with Latex :)

It is not clear why your code fragment has three consecutive identical tables. Anyway, in suggested solution is consider just one:

In code for above table is consider tblr table environment defined in the package tabularray which is nested in math environment. By this are eliminated all  in table body: Edit: In MWE is now considered tabularray version 2021N. \documentclass{article} \usepackage{tabularray} \begin{document} $\begin{tblr}{colspec = { l l X[l] X[1.1,l] X[l] }, vlines, hline{1,2,Z} = {1pt}, % <--- colsep=4pt, } t & y & y in terms of y_0\ [=(-\frac{1}{2})g\tau^2] & in terms of successive intervals~ & Ratio of distances traversed~ \\ 0 & 0 & 0 & & \\ \tau & -(\frac{1}{2}) g \tau^2 & y_0 & y_0 & 1 \\ 2\tau & -4(\frac{1}{2}) g \tau^2 & 4y_0 & 3y_0 & 3 \\ 3\tau & -9(\frac{1}{2}) g \tau^2 & 9y_0 & 5y_0 & 5 \\ 4\tau & -16(\frac{1}{2}) g \tau^2 & 16y_0 &7y_0 & 7 \\ 5\tau & -25(\frac{1}{2}) g \tau^2 & 25y_0 & 9y_0 & 9 \\ 6\tau & -36(\frac{1}{2}) g \tau^2 & 36y_0 & 11y_0 & 11 \\ \end{tblr}$ \end{document}  • Nice solution. // There are 3 tables, because the OP wanted to show some of his or her attempts, which got edited into one big source-code later. Jul 19 at 7:06 • @MS-SPO, thank you for compliment. About three tables: it is still not clear to me. But OP have two possibilities: repeat table body three times (in one big table) or repeat the same table three times. Jul 19 at 7:36 To create multiple lines in the column header, the \thead command from the makecell package can be used. \documentclass{article} \usepackage{makecell} \usepackage{amsmath} \renewcommand\theadalign{tl} %for top left alignment \begin{document} \begin{tabular}{|l|l|l|l|l|} \hline \thead{t$} & \thead{$y$} & \thead{$y$in terms of\\$y_0[=\left(-\frac 12\right) g\tau^2]$} & \thead{Distance \\traversed in \\successive \\intervals} & \thead{Ratio of \\distances \\traversed} \\ \hline$0$&$0$&$0$&$\:$&$\:$\\$ \tau$&$-\left(\frac 12\right) g \tau^2$&$y_0$&$y_0$&$1$\\$2 \tau$&$-4\left(\frac 12\right) g \tau^2$&$4y_0$&$3y_0$&$3$\\$3 \tau$&$-9\left(\frac 12\right) g \tau^2$&$9y_0$&$5y_0$&$5$\\$4\tau$&$-16\left(\frac 12\right) g \tau^2$&$16y_0$&$7y_0$&$7$\\$5 \tau$&$-25\left(\frac 12\right) g \tau^2$&$25y_0$&$9y_0$&$9$\\$6 \tau$&$-36\left(\frac 12\right) g \tau^2$&$36y_0$&$11y_0$&$11\$ \\
\hline
\end{tabular}
\end{document}