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I am having some trouble with a math question. If anyone can help that would be great.The question is:

The height (h) in centimeters of Shuntaro's pogo stick from the ground as he jumps on it since s seconds have passed, is roughly modelled by the equation h(s) = sin^(2)4s + 2sin4s + 1 a) how many times does Shuntaro hit the ground in the first 5 seconds b) how many centimeters is shuntaro off the ground after 2 seconds? c) During the first second, when is Shuntaro at 2.5 centimeters to two decimal places

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    Please search for how to use LaTeX formatting to make your question readable! – 2017-01-07
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    You seem to be mixing the use of $s$'s and $x$'s. For part (b) the answer requires just plug-and-chug of the numbers. *The height of Shuntaro at time $s$ is $h(s)$*... so... the height at time $2$ is... As for the rest, what analytical tools do you have available to you. Can you use calculus? If not, do you at least have a few convenient points on the unit circle memorized? If the height is $2.5$ that implies something about $\sin^2(4s)$ and $2\sin(4s)$, what value should $\sin(4s)$ be? What value(s) of $s$ makes that possible? – 2017-01-07

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So I'm translating your question this way:

1) The height of Shuntaro on his pogostick is modeled by $$h(t) = sin^2(4t) +2sin(4t) +1$$

2) So we are looking for all values of $t$ such that $$sin^2(4t) + 2 sin(4t) = -1$$

3) So note first that $sin^2(4x)$ varies between 0 and 1; do you see why?

In the first two quadrants (for $0 \leq t \leq \pi$ the function is non-negative. In the next two quadrants $sin^2(t)$ starts going back up towards 1 and then back down to 0, whereas $sin(4t)$ heads towards -1, and then back up to 0.

Thus the only place where the left hand side of 2), above, can equal $-1$ is when $4t =\frac{3 \pi}{2}$. But because $sin$ is periodic, we actually need all times that $4t = \frac{3 \pi}{2} + 2n\pi$ where $n$ is any integer. This means that $t = \frac{1}{8}(4 \pi n + 3 \pi)$. This next part needs to be done numerically, but in the end you will find that Shuntaro hits the ground 3 times in the first 5 seconds.

Can you think how you would solve the next two parts? Do you know what it means to say $h(t) = ...$ etc? It means if you plug a number in for $t$ on the right hand side, the function $h$ spits back out the height. You can do that part. Then for part c, the question is saying that at some time $t$, the left hand side is equal to 2.5 cm. Can you use what I said above to find those times?