Question

7. A rock is thrown with a height equation of h=-1612 20t 5 (where h is the height of the rock in feet at any given time of t in seconds). Will it reach a height of 30 feet? Explain your answer

Answer 1

Since h(max) is less than 30 ft, it will never reach a height of 30ft

The maximum height it can reach is 11.25ft.

Explanation:

Given;

The height equation of the rock;

h = -16t^2 +20t +5

To determine whether it would reach 30 ft, we need to find its maximum height. Which is at;

dh/dt = 0

dh/dt = -32t +20 = 0

At Maximum height.

t = 20/32

We then substitute into the height equation.

h(max) = -16(20/32)^2 + 20(20/32) +5

h(max) = 11.25 ft

Since h(max) is less than 30 ft, it will never reach a height of 30ft

The maximum height it can reach is 11.25ft.

Answer 2

There are no real solution to the equation
`h = -16t^2 +20t +5` when h = 35 ft.

Explanation:

Here we have the equation for the height of the rock in ft given as

`h = -16t^2 +20t +5`

For the rock to reach 30 ft we must have

`30 = -16t^2 +20t +5`

That is
`0 = -16t^2 +20t +5 - 30 = -16t^2 +20t -25`

`0 = -16t^2 +20t -25` or

`16t^2 -20t +25 = 0`

From which it is observed that since the root of the equation is given by the quadratic formula

b² should be greater than 4·a·c however

(-20)²
`\\geqslant` (4×16×25)

Hence the equation has only imaginary roots.