Question

3) Cory is standing on a cliff that is 48 feet above the ground. He throws a rock in to the air. The height, h, of the rock

after t seconds is given by h(t) = -2t+4t +48.


a) What is the initial height?


b) How many seconds does it take to reach the maximum height?


c) What is the rock's maximum height?


d) How many seconds does it take to for the rock to land on the ground?

Answer 1

a) The initial height of the rock is of 48 feet.

b) It takes 1 seconds for the rock to reach maximum height.

c) The rock's maximum height is 50 feet.

d) It takes 6 seconds for the rock to land on the ground.

Explanation:

Vertex of a quadratic function:

Suppose we have a quadratic function in the following format:

`f(x) = ax^(2) + bx + c`

It's vertex is the point
`(x_(v), f(x_(v))`

In which

`x_(v) = -(b)/(2a)`

If a<0, the vertex is a maximum point, that is, the maximum value happens at
`x_(v)`, and it's value is
`f(x_(v))`

Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

`ax^(2) + bx + c, a\\eq0`.

This polynomial has roots
`x_(1), x_(2)` such that
`ax^(2) + bx + c = a(x - x_(1))*(x - x_(2))`, given by the following formulas:

`x_(1) = (-b + √(\bigtriangleup))/(2*a)`

`x_(2) = (-b - √(\bigtriangleup))/(2*a)`

`\bigtriangleup = b^(2) - 4ac`

In this question:

The height of the ball after t seconds, in feet, is given by:

`h(t) = -2t^2 + 4t + 48`

Which is a quadratic equation with
`a = -2, b = 4, c = 48`

a) What is the initial height?

This is
`h(0) = 48`. So the initial height of the rock is of 48 feet.

b) How many seconds does it take to reach the maximum height?

This is the value of t at the vertex. So

`t_(v) = -(b)/(2a) = -(4)/(2(-2)) = 1`

It takes 1 seconds for the rock to reach maximum height.

c) What is the rock's maximum height?

This is the height of the ball after 1 second. So

`h(1) = -2(1)^2 + 4(1) + 48 = 50`

The rock's maximum height is 50 feet.

d) How many seconds does it take to for the rock to land on the ground?

This is t for which
`h(t) = 0`, so we solve the quadratic equation.

`\bigtriangleup = (4)^2-4(-2)(48) = 400`

`t_(1) = (-4 + √(400))/(2*(-2)) = -4`

`t_(2) = (-4 - √(400))/(2*(-2)) = 6`

Time is a positive measures, so it takes 6 seconds for the rock to land on the ground.