Question

A ball is thrown from an initial height of 7 feet with an initial upward velocity of 24 fts. The ball's height h (in feet) after seconds is given by the following.
h=7+24t-16t^2
Find all values of t for which the ball's height is 11 feet.
Round your answer(s) to the nearest hundredth.

Answer 1

Answer:

The ball's height is 11 feet at t = 0.19s and t = 1.31s

Step-by-step explanation:

The ball's height h (in feet) after seconds is given by the following function:

$h(t) = 7 + 24t - 16t^{2}$

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

This is t when $h(t) = 11$.

We will need to solve a quadratic equation for this.

Solving a quadratic equation

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$.

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

$x_{1} = \frac{-b + \sqrt{\bigtriangleup}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\bigtriangleup}}{2*a}$

$\bigtriangleup = b^{2} - 4ac$

In this problem:

$h(t) = 11$

$7 + 24t - 16t^{2} = 11$

$16t^{2} - 24t + 4 = 0$

Simplifying by 4

$4t^{2} - 6t + 1 = 0$

So $a = 4, b = -6, c = 1$

$\bigtriangleup = (-6)^{2} - 4*(4)*1 = 20$

$x_{1} = \frac{-(-6) + \sqrt{20}}{2*4} = 1.31$

$x_{2} = \frac{-(-6) - \sqrt{20}}{2*4} = 0.19$

The ball's height is 11 feet at t = 0.19s and t = 1.31s