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{\bf Question}

A particle is thrown vertically upwards from the surface of the
earth with a speed 1ms$^{-1}$.

(You may assume the gravitational field is constant with g =
9,8ms$^{-2}$)
\begin{description}
\item[(i)] How long does it take for it to return to earth?
\item[(ii)] What is its maximum height?
\item[(iii)] What is its maximum speed?
\end{description}

\vspace{.25in}

{\bf Answer}

\setlength{\unitlength}{.25in}
\begin{picture}(4,4.5)
\put(0,0){\line(1,0){4}}

\put(1,0){\vector(0,1){3}}

\put(1,3.2){$y$}

\put(1.9,2.4){$\bullet$}

\put(2,2.5){\vector(0,-1){1}}

\put(1.9,1.1){$mg$}

\end{picture}

\bigskip

Newton's 2nd Law: $ m\ddot{y} = -mg \Rightarrow  \ddot y = -g$

Therefore $y = vt - \frac{1}{2}gt^2$

Initially $\dot y = v = 1{\rm ms}^{-1}$

\begin{description}
\item[(i)]
The particle returns to the ground when $y = 0.$

$\ds \begin{array}{cccccc} vt - \frac{1}{2}gt^2 = 0 & \Rightarrow
& t(v - \frac{1}{2}gt) = 0 \\ & \Rightarrow & t = 0 {\rm \ \ \ or
\ \ }& v - \frac{1}{2} gt  & = &  0 \\  & & & \frac{1}{2} gt & = &
v
\\ & & & t & = & \frac{2v}{g} \\ ({\rm since}\  v = 1,\  {\rm and}\  g =
9.8)& & & t & = & \frac{2 \times 1}{9.8} \\ & & & & \approx & 0.2
s
\end{array}$

\item[(ii)]
The maximum height occurs when there is no upward speed.

Thus $\ds \dot y = v - gt = 0 \Rightarrow t = \frac{v}{g} =
\frac{1}{9.8} s$

\item[(iii)]
The maximum speed occurs at $y = 0,$ and is therefore 1ms$^{-1}$

\end{description}



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