Question

5. The height of a crab, h(t) in metres, dropped by a seagull to the rocks below is given by the

function, h(t) = -5t² + 20 where t is the time, in seconds, after its release.

a) What is the initial height of the crab before being dropped?

b) When does the crab hit the rocks?

Answer 1

Answer: (-5,0) & (4,0) are the zeros

Explanation:

According to the actual situation

When y = -5, h = -5<0

When t = 4, h(4), h = 5x4 + 20 = -20 + 20 = 0

Answer: (-5,0) & (4,0) are the zeros

Answer 2

Explanation:

a) To find the initial height of the crab before being dropped, we need to find the value of h(t) when t is 0 (at the initial time).

Plugging in t = 0 into the height function, we have:

h(0) = -5(0)^2 + 20

h(0) = 20

Therefore, the initial height of the crab before being dropped is 20 meters.

b) To find when the crab hits the rocks, we need to find the value of t when h(t) is 0 (height is 0).

Setting h(t) = 0 in the height function, we have:

-5t² + 20 = 0

Solving this quadratic equation, we can factor it as:

-5t² + 20 = 0

-5(t² - 4) = 0

Setting each factor equal to zero, we have:

t² - 4 = 0

Taking the square root of both sides, we get:

t = ± √4

t = ± 2

Since time cannot be negative in this context, the crab hits the rocks at t = 2 seconds.

Therefore, the crab hits the rocks 2 seconds after being dropped.