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Tower Escape is one of the most popular thrill rides at Mountain Adventure amusement park. It lifts riders to the top of a tower and drops them from a height of 200 feet. The riders fall to a height of 75 feet, at which point magnetic brakes slow them down.
Which equation can you use to find how many seconds the riders fall before slowing down?
If an object falls from s feet above the ground, the object's height in feet, h, after t seconds can be modeled by the formula h=

16t2+s.
To the nearest tenth of a second, how long do the riders fall before slowing down?
seconds

1 answer

To determine how many seconds the riders fall before slowing down, we need to use the given height formula:

\[ h = -16t^2 + s \]

In this formula, \( h \) is the height after \( t \) seconds, and \( s \) is the initial height from which the riders fall.

  1. Identify the variables:

    • The starting height \( s \) is 200 feet (the height of the tower).
    • The riders fall to a height of 75 feet before the brakes are activated.
  2. Set up the equation: We want to find \( t \) when the height \( h \) is 75 feet. So, we can set up the equation:

\[ 75 = -16t^2 + 200 \]

  1. Rearranging the equation: Subtract 200 from both sides:

\[ 75 - 200 = -16t^2 \]

\[ -125 = -16t^2 \]

  1. Divide by -1 to simplify:

\[ 125 = 16t^2 \]

  1. Solve for \( t^2 \):

\[ t^2 = \frac{125}{16} \]

  1. Taking the square root:

\[ t = \sqrt{\frac{125}{16}} \]

Calculating that:

\[ t = \frac{\sqrt{125}}{4} = \frac{5\sqrt{5}}{4} \]

  1. Now, compute \( \sqrt{125} \):

Since \( \sqrt{125} = 11.1803 \) (approximately),

\[ t \approx \frac{11.1803}{4} \approx 2.795 \]

  1. Round to the nearest tenth:

\[ t \approx 2.8 \]

Therefore, the riders fall for approximately 2.8 seconds before slowing down.