Question

A rocket is fired from the deck of a barge in Lake Ontario at a Canada Day fireworks display. The path of the rocket can be approximately modeled by the equation ℎ=−3t^2+24t+57, where h is the height, in metres, of the rocket above the water and t is the time, in seconds, after it was launched. - When did the rocket reach its maximum height and what was the maximum height achieved? - How long was the rocket above the water after it was launched?

Answer 1

Answer:

Step-by-step explanation:

ℎ = −3t² + 24t + 57

The graph/parabola of given quadratic equation opens down, y-intercept of the parabola is point (0, 57), that means the height of the deck is 57 m.

- When did the rocket reach its maximum height?

The maximum height is the vertex of a parabola with coordinate (t, h). The t-value is $\frac{-b}{2a}$ , which is $\frac{-24}{2*(-3)}$ = 4 seconds. Therefore, the rocket reaches its maximum height in 4 seconds after being fired.

- What was the maximum height achieved?

Let substitute 4 into the function to find the maximum height.

h(t) = - 3 × 4² + 24 × 4 + 57 = 105 m

The maximum height is 105 m.

- How long was the rocket above the water after it was launched?

The rocket hits the water when the height is 0. Therefore, we have to solve the quadratic equation  −3t² + 24t + 57 = 0 ===>  − t² + 8t + 19 = 0

$t_{12}$ = [ - 8 ± √(64 + 76)] / (- 2) = ( - 8 ± √140) / (- 2)

$t_{1}$ = ( - 8 + √140) / (- 2) ≈ - 1.916 seconds is not a solution of the problem.

$t_{2}$ = ( - 8 - √140) / (- 2) ≈ 9.916 seconds.

The rocket was above the water 9.916 seconds.

Answer 2

Given :  The path of the rocket can be approximately modeled by the equation ℎ=−3t^2+24t+57, where h is the height, in metres, of the rocket above the water and t is the time, in seconds, after it was launched.

To find :   When did the rocket reach its maximum height and what was the maximum height achieved

Solution:

ℎ=−3t²+24t+57

dh/dt =  -6t + 24  

putting

dh/dt = 0

=>  -6t + 24   = 0

=> t = 4

dh/dt =  -6t + 24  

d²h/dt²  = - 6  < 0

=> at t = 4  , h will be maximum

rocket reach its maximum height  at t = 4 sec

h = −3t²+24t+57

= - 3 * 4² + 24*4 + 57

= -48 + 96 + 57

= 105

maximum height achieved = 105  m

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