# Math Help - Stuck with hard question

1. ## Stuck with hard question

Here is the question,

Two bungy jump operators are conducting an experiment to estimate the length of the latex rubber required, when full extended, for a safe vertical jump off the bridge,

OP1 drops a boulder from the bridge, uses a stopwatch to time how long it takes before he sees the splash in the river below. This takes 4.2 seconds.

OP2 does the same, uses a stopwatch but he times how long it takes before he hears the splash. This takes 4.5 seconds.

The relationship between the distance the boulder has fallen and the time take is d=4.9t^2. The speed of sound in air is 344m/s.

Their final estimate of the safe distance is the mean of their estimates, less a 5m saftey margin.

Can anyone help me with this please, apparently it is a quadratic that is solved by completing the square.

2. ## Re: Stuck with hard question

Originally Posted by henryNZ
Here is the question,

Two bungy jump operators are conducting an experiment to estimate the length of the latex rubber required, when full extended, for a safe vertical jump off the bridge,

OP1 drops a boulder from the bridge, uses a stopwatch to time how long it takes before he sees the splash in the river below. This takes 4.2 seconds.

OP2 does the same, uses a stopwatch but he times how long it takes before he hears the splash. This takes 4.5 seconds.

The relationship between the distance the boulder has fallen and the time take is d=4.9t^2. The speed of sound in air is 344m/s.

Their final estimate of the safe distance is the mean of their estimates, less a 5m saftey margin.

Can anyone help me with this please, apparently it is a quadratic that is solved by completing the square.
The speed of light in air is so fast that you may as well consider it to be isntantaneous.

So $t_1=4.2s$ and thus $\hat{d}_1=4.9 (4.2)^2=86.436m$

To find $\hat{d}_2$ we note that $t_2$ is the time for the boulder to reach the river plus the time it takes for the sound to travel back the same distance.

$t_2=4.5=\sqrt{\dfrac{\hat{d}_2}{4.9}}+\dfrac{\hat {d}_2}{344}$

$4.5 - \dfrac{\hat{d}_2}{344}=\sqrt{\dfrac{\hat{d}_2}{4.9 }}$

$\left(4.5 - \dfrac{\hat{d}_2}{344}\right)^2=\dfrac{\hat{d}_2}{ 4.9}$

$8.45051\times10^{-6} \hat{d}_2^2-0.0261628\hat{d}_2+20.25=0.204082\hat{d}_2$

$8.45051\times10^{-6} \hat{d}_2^2-0.230244\hat{d}_2+20.25=0$

just solve this using the quadratic formula and reject the nonsensical root that appears because we squared the original equation.

Doing this we find $\hat{d}_2=88.2358m$

I leave it to you to find the mean and subtract off the safety margin.