A ball was thrown up from the top of a cliff as shown below. The height of the ball, h metre, from the sea level is measured, t seconds after leaving the cliff. The table below shows the height, h, reached by the ball after time t.

1. Draw and label the axes using the required scales.

Horizontal axis: $t$ in seconds, using $2\text{ cm}$ to represent $0.5\text{ s}$.

Vertical axis: $h$ in metres, using $1\text{ cm}$ to represent $2\text{ m}$.

The vertical axis should include values from below $0$ up to at least about $28\text{ m}$, since the table includes $h=-3.12$ and a maximum height slightly above $26\text{ m}$.

2. Plot all given coordinate points accurately.

Plot the points:

3. Join the plotted points appropriately.

Join the plotted points with a smooth curve, not with straight line segments, since the data represents the curved path of the ball.

1. Identify the initial height of the ball, which is the height of the cliff.

At $t=0$, the ball has just left the top of the cliff.

From the graph/table, when $t=0$, $h=25$.

Hence, the height of the cliff is $25$ m.

1. Identify where the ball reaches sea level on the graph.

Sea level means the height from sea level is $h=0$.

On the graph of $h$ against $t$, read the value of $t$ where the curve crosses the horizontal axis $h=0$.

This gives approximately
$t \approx 3.7.$

1. Draw the tangent at $t=2$ on the graph.

On the graph, locate the point on the curve where $t=2$. From the table/graph, this point is approximately $(2,21)$. Draw a straight line that just touches the curve at this point and follows the local direction of the curve.

2. Calculate the gradient of the tangent.

Choose two clear points on the tangent and calculate
$\text{gradient}=\frac{\text{vertical change}}{\text{horizontal change}}.$
From a correctly drawn tangent, the gradient is approximately
$\frac{-16}{2}=-8.$

1. Interpret the gradient of the height-time graph in the context of the motion.

For a graph of $h$ against $t$,
$\text{gradient}=\frac{\text{change in height}}{\text{change in time}}.$
Hence the gradient represents the rate of change of the ball's height with time, i.e. its vertical velocity at that instant.

1. Draw the straight line representing the toy plane on the existing graph.

For the line $h=20+7t$, choose two convenient points:

When $t=0$, $h=20$.

When $t=1$, $h=27$.

Plot $(0,20)$ and $(1,27)$, then join them with a straight line.

2. Read the collision time from the intersection of the two graphs.

The toy plane collides with the ball when both have the same height at the same time.

This is the point where the line $h=20+7t$ intersects the curve of the ball.

From the graph, the intersection occurs at about

$t \approx 0.9.$

1. Recognise what $A$ and $B$ represent in the given form.

The equation is

$h=A-3(t-B)^2.$

Since $-3(t-B)^2 \leq 0$, the maximum value of $h$ occurs when

$t-B=0,$

so

$t=B \quad \text{and} \quad h=A.$

Hence the vertex, or highest point, of the graph is $(B,A)$.

2. Read the maximum point from the graph.

From the graph, the highest point of the curve is approximately

$(0.7,\ 26.3).$

Therefore,

$B \approx 0.7, \qquad A \approx 26.3.$