# Thread: Word problems - 2 problems - trig gr. 12 need immediate help ty!

1. ## Word problems - 2 problems - trig gr. 12 need immediate help ty!

1st: A pendulum is connected to a rope 3 m long, which is connected to a ceiling 4 m high. The angle between its widest swing and vertical hanging position is . If the pendulum swings out to its widest position in 2 seconds, model the height of the pendulum from the ground using a cosine function, considering vertical to be t = 0.

so far, i think i got that the period is 4 and the minimum is 1. i can't find anything else. please help.

2nd:
Tonight a full moon is out. The percentage of moon that is visible, v, according to the number of days from now, d, is modelled by the equation .
a) How many days is there until the new moon (no moon visible)?
b) Including tonight, how many full moons will there be in the next 100 days?
c) How many days will it be until there are five full moons, including tonights?
d) Without using trigonometric ratios, write a simple equation that tells how many days away the next x full moons are. Why can the equation be so simple and consistent?

Verify my work please? let me know if i'm right or wrong. 2. ## Re: Word problems - 2 problems - trig gr. 12 need immediate help ty!

1st: A pendulum is connected to a rope 3 m long, which is connected to a ceiling 4 m high. The angle between its widest swing and vertical hanging position is . If the pendulum swings out to its widest position in 2 seconds, model the height of the pendulum from the ground using a cosine function, considering vertical to be t = 0.

so far, i think i got that the period is 4 and the minimum is 1. i can't find anything else. please help.

Have you drawn a diagram Can you use right-angled triangle trig to find the highest vertical position reached by the pendulum? This will help you find the amplitude.

4. ## Re: Word problems - 2 problems - trig gr. 12 need immediate help ty!

(1) range of the pendulum's height above the ground is $1 \le h(t) \le 2.5$, where $h(0) = 1$, $h(2) = 2.5$ and $h(4) = 1$

Model for pendulum bob height vs time graph is an inverted cosine curve, $h(t) = D - A\cos(Bt)$

Amplitude is $A = \dfrac{2.5-1}{2} = 0.75$

Midline is $D = \dfrac{1+2.5}{2} = 1.75$

The period is $T=4 \implies B = \dfrac{2\pi}{T} = \dfrac{\pi}{2}$

$h(t) = 1.75 - 0.75\cos\left(\dfrac{\pi}{2} \cdot t\right)$

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help needed, immediately, please help, thank you 