Originally Posted by

**Soroban** Hello, magentarita!

I will assume that: .$\displaystyle 0.157 \:\approx\:\frac{\pi}{20}$

**No, it's 0.157 minus pi/2...not approximately. How does knowing this now change the answers given?**

Can't you set up the equation?

The question is: When is $\displaystyle h = 125$?

We have: .$\displaystyle 125\sin\left(\frac{\pi}{20}t - \frac{\pi}{2}\right) + 125 \:=\:125 \quad\Rightarrow\quad \sin\left(\frac{\pi}{20}t - \frac{\pi}{2}\right) \:=\:0$

Then: .$\displaystyle \frac{\pi}{20}t - \frac{\pi}{2} \;=\;\{0,\pi,2\pi,3\pi,\hdots\} \quad\Rightarrow\quad \frac{\pi}{20}t \;=\;\left\{\frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2},\frac{7\pi}{2},\hdots\right\} $

Hence: .$\displaystyle t \;=\;\{10, 30, 50, 70, \hdots \}$

Answers: . $\displaystyle \boxed{t \;=\;10\text{ sec},\;30\text{ sec}}$

The rider is above 125 feet on the interval: $\displaystyle t \in (10,30)$