h=-5t to the 2nd power+30t+10

Question

a model rocket is launched from the roof of a building. its flight path is modeled by h+-5t to the second power+30t+10 where h is the height of the rocket above the ground in meters and t is the time after the launch in seconds so the qestions are what is the initial height of hte rocket at hte time time of launch? what is the maximum height of the rocket? when does the rocket hit the ground?

Answer 1

Initial height:

h(0) = 10 ft

Time to max height:

Find the vertex of the height-vs.-time parabola.

t = -30/[2(-5)] sec = 3 sec

 

Now plug t = 3 into h(t) to get h_max.

 

Time to hit ground:

 

Set h(t) = 0 and solve for t.

 

-5t² + 30t + 10 = 0

 

t² - 6t - 2 = 0

 

Solve this quadratic equation for t and get

 

t ≅ -0.32 or 6.32 sec.

 

Since t starts at t = 0, discard the negative solution, and the answer is about 6.32 sec.

Answer 2

The initial height is the y-intercept, when t = 0.

 

h(0) = -5(0)² + 30(0) + 10

h(0) = 10 feet

 

To find the max height, we first find the time it reaches it's maximum height.

 

t = -b/2a = -30/(2(-5)) = -30/-10 = 3 seconds

 

h(3) = -5(3)² + 30(3) + 10

h(3) = -5(9) + 90 + 10

h(3) = 55 feet

 

it reaches its high point halfway through flight, so it hits the ground in 2(3) = 6 seconds.