The height (h) in centimetres 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(4x) + 2sin^2(4x) + 1
a) How many times does Shuntaro hit the ground in the first 5 seconds?
b) How many centimetres is Shuntaro off the ground after 2 seconds?
c) During the first second, when is Shuntaro at 2.5 centimetres to two decimal places?
really? h(s) depends only on x? and why do you have sin^2(4x) twice?
and that's some puny pogo stick, if it never gets more than 4 cm off the ground!
(a) sin^2(4x) = 1/2 (1 - cos(8x)), so it has a period of π/4. It hits the x-axis at the end of each period.
(b) just find h(2)
(c) solve h(s) = 2.5
so fix your function and use the hints above.
Please check for typos.
Do you mean h(s) of h(x)?
Do you really mean 3sin^2(4x) + 1 ?
also, h(s) never is equal to zero. The pogo stick never touches the ground?
As written it is never less than 1 cm above ground by the way sin^2 is always +
To answer these questions, let's break down the given equation and then use it to find the answers.
The equation provided is h(s) = sin^2(4s) + 2sin^2(4s) + 1.
a) To determine how many times Shuntaro hits the ground in the first 5 seconds, we need to find the number of "zero crossings" or where the height equals zero.
We can set the equation h(s) = 0 and solve for s.
0 = sin^2(4s) + 2sin^2(4s) + 1.
By substituting y = sin^2(4s) and expanding, we get:
0 = y + 2y + 1
0 = 3y + 1
3y = -1
y = -1/3.
Since y is the square of the sine function and must be between 0 and 1, y = -1/3 is not possible. Therefore, there are no zero crossings in the first 5 seconds, which means Shuntaro does not hit the ground within that time interval.
b) To find Shuntaro's height after 2 seconds, we can substitute s = 2 into the equation h(s) = sin^2(4s) + 2sin^2(4s) + 1:
h(2) = sin^2(4 * 2) + 2sin^2(4 * 2) + 1.
Calculating this expression:
h(2) = sin^2(8) + 2sin^2(8) + 1.
Using a scientific calculator or computer program, we can evaluate sin^2(8) and 2sin^2(8):
h(2) ≈ 0.6953 + 2(0.6953) + 1
h(2) ≈ 0.6953 + 1.3906 + 1
h(2) ≈ 3.086.
Therefore, Shuntaro is approximately 3.086 centimeters off the ground after 2 seconds.
c) To find when Shuntaro is at 2.5 centimeters within the first second, we need to solve the equation h(s) = 2.5 for s. However, this equation cannot be solved using elementary algebraic techniques since it involves trigonometric functions. Therefore, we need to use numerical methods, such as graphical or numerical approximation.
One way to estimate when h(s) ≈ 2.5 is to plot the function h(s) = sin^2(4s) + 2sin^2(4s) + 1 and visually inspect the graph for when it intersects the height of 2.5. Alternatively, we can use iterative methods like the bisection method or Newton's method to find a numerical approximation.
Unfortunately, without further information or the use of numerical methods, we cannot determine the exact time when Shuntaro is at 2.5 centimeters within the first second.