WAEC 2014 · Paper 2 · Q9

(a)
An object was thrown vertically upwards from the top of a cliff and its height, $y$ metres, above sea level after $t$ seconds is given by $y = -16t^2 + 64t + 5$ . Copy and complete the table of values for $0 \le t \le 4.0$ .

$t$ 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

$y$ 5 65 53

Model answer

$t$ 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

$y$ 5 33 53 65 69 65 53 33 5

For example $t = 0.5$ : $-4 + 32 + 5 = 33$ , and $t = 2.0$ : $-64 + 128 + 5 = 69$ . The values are symmetrical about $t = 2$ .
(b)
Using scales of 2 cm to 0.5 seconds on the $t$ -axis and 2 cm to 10 m on the $y$ -axis, draw the graph of $y = -16t^2 + 64t + 5$ for $0 \le t \le 4.0$ .
Model answer
Plot every point from the table, then join them with one smooth curve (not straight lines between points). Scale: 2 cm to 0.5 s across, 2 cm to 10 m up. The path is symmetrical about $t = 2$ .

For (c): (i) at $t = 1.75$ s the height is about 68 m; (ii) the line $y = 50$ meets the curve at $t \approx 0.9$ s and $3.1$ s; (iii) the maximum height is 69 m.
(c)
Use the graph to find the: (i) height reached when $t = 1.75$ seconds; (ii) times the object was at a height of $50\text{ m}$ ; (iii) maximum height reached.
Separate values with commas, e.g. 3, −2

$t$	0.0	0.5	1.0	1.5	2.0	2.5	3.0	3.5	4.0
$y$	5	33	53	65	69	65	53	33	5

The height curve with the line y = 50.

Worked solution (try it first)

Substitute each $t$ into $y = -16t^2 + 64t + 5$ .
For example, $t = 0.5$ gives $-4 + 32 + 5 = 33$ and $t = 2$ gives $-64 + 128 + 5 = 69$ .
$t$ 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

$y$ 5 33 53 65 69 65 53 33 5

$t$	0.0	0.5	1.0	1.5	2.0	2.5	3.0	3.5	4.0
$y$	5	33	53	65	69	65	53	33	5

With 2 cm to 0.5 s on the $t$ -axis and 2 cm to 10 m on the $y$ -axis, plot the points and join them with a smooth curve shaped like an upside-down U.

Go up from $t = 1.75$ to the curve and across: the height is about $68\text{ m}$ .

The highest point of the curve is at $t = 2$ : the maximum height is $69\text{ m}$ .
(Check: $\frac{dy}{dt} = -32t + 64 = 0$ at $t = 2$ .)