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Stunt Double A movie stunt double jumps from the top of a
building 50 feet above the ground onto a pad on the ground
below. The stunt double jumps with an initial vertical velocity
of 10 feet per second.
a. Write and graph a function that models the height h (in feet)
of the stunt double t seconds after she jumps.
b. How long does it take the stunt double to reach the ground?

Respuesta :

znk

Answer:

2.1 s

Step-by-step explanation:

1. Write the equation

The formula for vertical displacement is

y =  h + v₀t - ½gt²

h = 50     ft

v₀ = 10     ft/s

g = 32.2 ft/s²

Insert the values

y = 50 + 10t - ½ × 32.2t²

y = 50 + 10t - 16.1t²

The graph below shows the height as a function of time.

2. Solve the equation

a = -16.1; b = 10; c = 50

[tex]\begin{array}{rcl}x & = & \dfrac{-b\pm\sqrt{b^2 - 4ac}}{2a}\\\\& = & \dfrac{-10\pm\sqrt{10^2-4(-16.1) \times 50}}{2(-16.1)}\\\\& = & \dfrac{-10\pm\sqrt{100 - (-3220)}}{-32.2}\\\\& = & \dfrac{-10\pm\sqrt{100 + 3220}}{-32.2}\\\\& = & \dfrac{-10\pm\sqrt{3300}}{-32.2}\\\\& = & \dfrac{-10\pm 57.62}{-32.2}\\\\\end{array}[/tex]

[tex]\begin{array}{rcl}x = \dfrac{-10 + 57.62}{-32.2}& \qquad & x = \dfrac{-10 -57.62}{-32.2}\\\\x = \dfrac{47.62}{-32.2}& \qquad & x = \dfrac{-67.62}{-32.2}\\\\x = -1.479& \qquad & x =\textbf{ 2.1 s}\\\\\end{array}\\\text{It takes the stunt double $\large \boxed{\textbf{2.1 s}}$ to reach the ground.}[/tex]

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