Question

12. Elmo is standing on a parade float tossing candy out to everyone. The candy is thrown with a starting velocity of 10

Answer 1

Answer:

c. 0.71

Step-by-step explanation:

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 $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$, given by the following formulas:

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

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

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

The height of the candy after t seconds is given by the following equation:

$h(t) = -16t^2 + 10t + 3$

After how many seconds will the ball be 2 feet in the air?

This is t for which $h(t) = 2$

So

$h(t) = -16t^2 + 10t + 3$

$2 = -16t^2 + 10t + 3$

$-16t^2 + 10t + 1 = 0$

So

$a = -16, b = 10, c = 1$

$\Delta = b^{2} - 4ac = 10^2 -4(-16)(1) = 164$

$t_{1} = \frac{-10 + \sqrt{164}}{2*(-16)} = -0.09$

$t_{2} = \frac{-10 - \sqrt{164}}{2*(-16)} = 0.71$

Since time is a positive measure, the answer is 0.71, which is option c.