Question

Antonio's toy boat is bobbing in the water under a dock. The vertical distance HHH (in \text{cm}cmstart text, c, m, end text) between the dock and the top of the boat's mast ttt seconds after its first peak is modeled by the following function. Here, ttt is entered in radians. H(t) = {5}\cos\left({\dfrac{2\pi}{3}}t\right) - {35.5}H(t)=5cos( 3 2π ​ t)−35.5H, left parenthesis, t, right parenthesis, equals, 5, cosine, left parenthesis, start fraction, 2, pi, divided by, 3, end fraction, t, right parenthesis, minus, 35, point, 5 How long does it take the toy boat to bob down from its peak to a height of -35\text{ cm}−35 cmminus, 35, start text, space, c, m, end text? Round your final answer to the nearest tenth of a second.

Answer 1

Answer:

0.7 seconds

Step-by-step explanation:

Answer 2

Answer: Time t = 33.0 seconds

Step-by-step explanation:

Given that the vertical distance H between the dock and the top of the boat's mast t seconds after its first peak is modeled by the function

H(t) = 5cos( 2π/3 ​t) − 35.5H

Where the maximum vertical distance = 5

At the down position, H(t) = 0

5cos( 2π/3 ​t) − (35.5/100)H = 0

5cos( 2π/3 ​t) − 0.355 × 5 =0

5cos( 2π/3 ​t) − 0.1775 = 0

5cos( 2π/3 ​t) = 0.1775

cos( 2π/3 ​t) = 0.1775/5

cos( 2π/3 ​t) = 0.355

2π/3 ​t = cos^-1 (0.355)

2π/3 ​t = 69.2

2πt = 69.2 × 3

2πt = 207.6

t = 207.6/2π

t = 33.0 seconds