Luiza is jumping on a trampoline. H(t)H(t)

H(t)

H, left parenthesis, t, right parenthesis

models her distance above the ground (in

m\text{m}

m

start text, m, end text

)

tt

t

t

seconds after she starts jumping. Here,

tt

t

t

is entered in radians.

H(t)=−0.6cos⁡(2π2.5t)+1.5H(t) = -0.6\cos\left(\dfrac{2\pi}{2.5}t\ri ght) + 1.5

H(t)=−0.6cos(

2.5

2π



t)+1.5

H, left parenthesis, t, right parenthesis, equals, minus, 0, point, 6, cosine, left parenthesis, start fraction, 2, pi, divided by, 2, point, 5, end fraction, t, right parenthesis, plus, 1, point, 5

What is the second time when Luiza reaches a height of

1.2 m1.2\text{ m}

1.2 m

1, point, 2, start text, space, m, end text

?

Round your final answer to the nearest hundredth of a second.

The second time when Luiza reaches a height of 1.2 m = 2 08 s

Step-by-step explanation:

Complete Question

Luiza is jumping on a trampoline. Ht models her distance above the ground (in m) t seconds after she starts jumping. Here, the angle is entered in radians.

H(t) = -0.6 cos (2pi/2.5)t + 1.5.

What is the second time when Luiza reaches a height of 1.2 m? Round your final answer to the nearest hundredth of a second.

Solution

Luiza is jumping on trampolines and her height above the levelled ground at any time, t, is given as

H(t) = -0.6cos⁡(2π/2.5)t + 1.5

What is t when H = 1.2 m

1.2 = -0.6cos⁡(2π/2.5)t + 1.5

0.6cos⁡(2π/2.5)t = 1.2 - 1.5 = -0.3

Cos (2π/2.5)t = (0.3/0.6) = 0.5

Note that in radians,

Cos (π/3) = 0.5

This is the first time, the second time that cos θ = 0.5 is in the fourth quadrant,

Cos (5π/3) = 0.5

So,

Cos (2π/2.5)t = Cos (5π/3)

(2π/2.5)t = (5π/3)

(2/2.5) × t = (5/3)

t = (5/3) × (2.5/2) = 2.0833333 = 2.08 s to the neareast hundredth of a second.

Hope this Helps!!!

could you give more information?

abde or 1,2,4, 5 is the answer

step-by-step explanation: