Cambridge International AS and A Level Mathematics Pure Mathematics 1

P1^

5

Applications

EXAMPLE 5.17 A stone is projected vertically upwards with a speed of 30 m s−^1.
Its height, h m, above the ground after t seconds (t  6) is given by:
h = 30 t − 5 t^2.

(i) (^) Findd
d and
d
d
h
t
h
t
2
2.
(ii) Find the maximum height reached.
(iii) Sketch the graph of h against t.
SOLUTION
(i) ddht
= 30 − 10 t.
d
d
2
2
h
t

= −10.

(ii) For a stationary point, ddht

= 0

30 − 10 t = 0

⇒ 10(3 − t) = 0

⇒ t = 3.

d

d

2

2

h

t

< 0 ⇒ the stationary point is a maximum.

The maximum height is

h = 30(3) − 5(3)^2 = 45 m.

(iii)

Note

For a position–time graph, such as this one, the gradient, ddht, is the velocity and dd

2

2

h

t

is the acceleration.

K (metres)

0 3 6

45

W (seconds)

Figure 5.33