It is about trigonometry ?

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Answer 1 · 2018-05-01T23:10:27

Given that the height of the rider from the ground on the wheel after t th sec of his start from ground follows the following relation

$h = 60 (1 - cos (k t)$ $h=60(1-cos(kt)$ m.

Where $k$ $k$ is a constant and $t$ $t$ is the time in minute.

At $t = 0$ $t=0$

$h = 60 (1 - cos (k \cdot 0) = 0$ $h=60(1-cos(k*0)=0$ m.

(1)

Again

$h = 60 (1 - cos (k \cdot t))$ $h=60(1-cos(k*t))$ m.

The greatest or maximum height of the rider will be when cos(kt) is minimum i.e. $cos (k t) = - 1$ $cos(kt)=-1$

So $h_{max} = 60 (1 - (- 1)) = 120$ $h_"max"=60(1-(-1))=120$ m

Wheel undergoes one complete rotation in 30 min.

(2) So when $t = 30$ $t=30$ min, $h = 0$ $h=0$

$0 = (1 - cos (k \cdot 30)$ $0=(1-cos(k*30)$

$\Rightarrow cos (k \cdot 30) = 1 = cos (2 π)$ $=>cos(k*30)=1=cos(2pi)$

$\Rightarrow k = \frac{2 π}{30} = \frac{π}{15}$ $=>k=(2pi)/30=pi/15$

(3) when $h = 90$ $h=90$ m let time be $T$ $T$ min

We have

$90 = 60 (1 - cos (\frac{π}{15} \cdot T)$ $90=60(1-cos(pi/15*T)$

$\Rightarrow \frac{90}{60} = (1 - cos (\frac{π}{15} \cdot T)$ $=>90/60=(1-cos(pi/15*T)$

$\Rightarrow \frac{3}{2} = (1 - cos (\frac{π}{15} \cdot T)$ $=>3/2=(1-cos(pi/15*T)$

$\Rightarrow cos (\frac{π}{15} \cdot T) = 1 - \frac{3}{2} = - \frac{1}{2} = cos (\frac{2 π}{3})$ $=>cos(pi/15*T)=1-3/2=-1/2=cos((2pi)/3)$

So $T = \frac{2 π}{3} \cdot \frac{15}{π} = 10$ $T=(2pi)/3*15/pi=10$ min