36.0k views
4 votes
The function $h(x)=-16x^2 + 48x + 6$ models the height $h$ (in feet) of a ball ejected from a ball launcher after $x$ seconds. When does the ball reach a height of 30 feet? If necessary, round to the nearest thousandth.

A) $x = 0.750$
B) $x = 1.500$
C) $x = 2.250$
D) $x = 3.000$

1 Answer

5 votes

Final answer:

The ball reaches a height of 30 feet at approximately 0.750 seconds after launch, which corresponds to Option A.

Step-by-step explanation:

We're giving the quadratic function h(x) = -16x^2 + 48x + 6, which models the height of a ball after it's launched. To find out when the ball reaches a height of 30 feet, we set the equation equal to 30 and solve for x:

30 = -16x^2 + 48x + 6

Moving everything to one side gives us the quadratic equation:

0 = -16x^2 + 48x - 24

Dividing through by -16 to simplify, we get:

0 = x^2 - 3x + 1.5

This is a quadratic equation and we can use the quadratic formula, x = (-b ± √(b^2-4ac))/(2a), to find the roots. After applying the formula, we find:

x = (3 ± √(9 - 6))/(2)

Thus, there are two times where the ball reaches 30 feet, but we are looking for the earliest time when the height is ascending. Only one of these solutions will be a positive time near one of the options given and that is x = 0.750 seconds, which corresponds to Option A.

User Varghese
by
8.0k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories