## 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). W

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

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3 months 2022-02-16T09:22:11+00:00 2 Answers 0 views 0

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.

Step-by-step 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.

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

Step-by-step explanation:

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

[tex]h = -16t^2 +20t +5[/tex]

For the rock to reach 30 ft we must have

[tex]30 = -16t^2 +20t +5[/tex]

That is [tex]0 = -16t^2 +20t +5 – 30 = -16t^2 +20t -25[/tex]

[tex]0 = -16t^2 +20t -25[/tex] or

[tex]16t^2 -20t +25 = 0[/tex]

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[tex]\ngeqslant[/tex] (4×16×25)

Hence the equation has only imaginary roots.