Define speed. A ball is thrown up with a speed of $15m{s}^{-1}$. How high will it go before it begins to fall? $(g=9.8m{s}^{-2})$
[3 MARKS]

Definition: 1 Mark
Problem: 2 Marks

1. Speed is defined as the total distance traveled divided by the total time it took to travel that particular distance.
2. Given, initial speed of ball, $u=15m{s}^{-1}$
final speed of ball, $v=0$ (because the ball stops)
and acceleration due to gravity, $g=-9.8m{s}^{-2}$ (the ball is going up against the gravity, so the value of g is to be taken as negative).
Let the height be $h$.

From third equation of motion,
$v^2=u^2+2gh$
$\Rightarrow (0{)}^2=(15{)}^2+2\times (-9.8)\times h$
$\Rightarrow 0=225-19.6h$
$\Rightarrow h=\frac{225}{19.6}=\frac{225}{196}\times 10=(\frac{15}{14}{)}^2\times 10={1.07}^2\times 10=1.07\times 1.07\times 10=11.449$

$\Rightarrow h=11.45m$

Thus, the ball will go to a maximum height of 11.45 metres before it begins to fall.