Question

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?

Answer 1

2.1 s

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

`\begin{array}{rcl}x & = & (-b\pm√(b^2 - 4ac))/(2a)\\\\& = & (-10\pm√(10^2-4(-16.1) * 50))/(2(-16.1))\\\\& = & (-10\pm√(100 - (-3220)))/(-32.2)\\\\& = & (-10\pm√(100 + 3220))/(-32.2)\\\\& = & (-10\pm√(3300))/(-32.2)\\\\& = & (-10\pm 57.62)/(-32.2)\\\\\end{array}`

`\begin{array}{rcl}x = (-10 + 57.62)/(-32.2)& \qquad & x = (-10 -57.62)/(-32.2)\\\\x = (47.62)/(-32.2)& \qquad & x = (-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.}`