Question

Using a quadratic equation

A ball is thrown from an initial height of 1 meter with an initial upward velocity of 20 m/s. The ball's height h (in meters) after 1 seconds is given by the
following
h=1+20t-5t2
Find all values of 1 for which the ball's height is 11 meters.
Round your answer(s) to the nearest hundredth.
(If there is more than one answer or )

Answer 1

The ball height is 11 meters at t = 0.59 seconds and t = 3.41 seconds.

Explanation:

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 + √(\Delta))/(2*a)`

`x_(2) = (-b - √(\Delta))/(2*a)`

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

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

`h(t) = 1 + 20t - 5t^2`

Find all values of t for which the ball's height is 11 meters.

This is t for which
`h(t) = 11`. So

`h(t) = 1 + 20t - 5t^2`

`11 = 1 + 20t - 5t^2`

`5t^2 - 20t + 10 = 0`

Simplifying by 5

`t^2 - 4t + 2 = 0`

Which means that
`a = 1, b = -4, c = 2`

`\Delta = (-4)^2 - 4(1)(2) = 8`

`t_(1) = (-(-4) + √(8))/(2) = 3.41`

`t_(2) = (-(-4) - √(8))/(2) = 0.59`

The ball height is 11 meters at t = 0.59 seconds and t = 3.41 seconds.