Question

Richard throws a stone over a bridge into the river below. The stone's

height, in meters above the water, x seconds after it is thrown is modeled
by h () = = -5 (x – 8) (x + 4).
a.) How many seconds after being thrown will the stone meet its maximum
height?
b.) What is the maximum height of the stone?

Answer 1

a.) In 1 second stone reach its maximum height

b.) 45 m

Explanation:

v=

2

u

​

=u−gt

=u−

2

3

​

g

t=1.5s

u=30m/s

h=

2g

u

2

​

h=

2×10

30×30

​

=45m