Question

A golfer strikes a golf ball into the air from the ground. The height, h metres, of the ball

can be modelled by h = 32t – 4t2, where t is the time in seconds after it leaves the
ground.

i. On a sheet of graph paper, using a scale of 2 cm to represent 1 unit on the t -axis
and 1 cm to represent 5 units on the h -axis, draw the graph of h = 32t – 4t2
ii Use your graph to find he maximum height of the ball above the ground and the time

at which it occurs.
u dont have to draw the graph can u just find the values for me thanku

Answer 1

Explanation:

The golf ball reaches its maximum vertical height when its vertical velocity h' becomes zero. Taking the derivative of h and setting it to zero, we get

h' = 32 - 8t = 0

which gives us t = 4 seconds. So after 4 seconds, the ball will reach its maximum height and its value is

h = 32(4) - 4(4)^2

= 128 - 64

= 64

Answer 2

64 m, at time t = 4 sec

Explanation:

We must maximize the function h = 32t – 4t^2. One way in which to do this is to find the axis of symmetry, t = -b / [2a], and from that calculate the height:

a = -4, b = 32, c = 0

-32

and therefore the axis of symmetry is t = ---------------- = 4

2(-4)

The height, h, at time t = 4, is h(4) = 32(4) - 4(4)^2 = 64 m

The maximum height of the ball is 64 m, at time t = 4 sec