please help!!!5

Question

On a forward somersault five, nina's height (h) in metres above the water is approximately modelled by the h= -5t^2 + 7t + 4, (t) is time in second. after she leaved the diving board. Round in nearest tenth.

a) find ninas maximum height above the water.

b) how many seconds it take for her to hit the surface of the water.

c) how high is the diving board above the water?

Answer 1

(c)

The height h = -5t² + 7t + 4 where t is time.

To find the height of the diving board above the water,

we first differentiate the height function with respect to t and then equate it to Zero.

h' = -10t + 7.

-10t + 7 = 0

10t = 7

t = 0.7 sec.

Now height of the diving board above the water is h(0.7)

h = -5t² + 7t + 4

h(0.7) = -5(0.7)²+7(0.7)+4

h(0.7) = 6.45 mts.

Therefore the height of the diving board above the water is 6.45 mts.

Answer 2

(a)

The height of the Nina h = -5t² + 7t + 4 where t is time.

To find the maximum height of the Nina,

we first differentiate the height function with respect to t and then equate it to Zero.

h' = -10t + 7.

-10t + 7 = 0

10t = 7

t = 0.7 sec.

Now Maximum height of Nina is h(0.7)

h = -5t² + 7t + 4

h(0.7) = -5(0.7)²+7(0.7)+4

h(0.7) = 6.45 mts.

Therefore the maximum height above the water is 6.45 mts.

Answer 3

(b)

The height of the Nina above the water h = -5t² + 7t + 4 where t is time.

Now we equate h = 0 then

-5t² + 7t + 4 = 0

5t² - 7t - 4 = 0

Find the roots of the Quadratic Equation.

= (7±√129)/10

= 1.835 and -0.45

Therefore if 5t² - 7t - 4 = 0 then t = 1.835 or -0.45

Since time cannot be a negative value, then t = 1.835 sec.

Therefore time taken to reach the surface of the water is 1.8sec.