MATH SOLVE

4 months ago

Q:
# A model rocket is launched from the ground with an initial velocity of 160 ft/sec. how long will it take the rocket to reach its maximum height? Show all work in the space provided. Assume the model rocket’s parachute failed to deploy and the rocket fell back to the ground. How long would it take the rocket to return to Earth from the time it was launched? Show all work in the space provided.

Accepted Solution

A:

The formula we need is

[tex]h(t)=-16t^2+v_0 t+h_0[/tex], where v₀ is the starting velocity and h₀ is the initial height. Using the velocity and starting height from our problem we have

[tex]h(t)=-16t^2+160t+0[/tex]. The path of this rocket will be a downward facing parabola, so there will be a maximum. This maximum will be at the vertex of the graph. To find the vertex we start out with [tex]x= \frac{-b}{2a}[/tex], which in our case is [tex]x= \frac{-160}{2(-16)}= \frac{-160}{-32} = 5[/tex]. It will take 5 seconds for the rocket to reach its maximum height. Plugging this back into our formula gives us

[tex]h(5)=-16(5^2)+160(5)+0 \\=-16(25)+800 \\=-400+800=400[/tex]

The rocket's maximum height is 400 feet.

We set our formula equal to zero to find the time it takes to hit the ground, then we factor:

[tex]0=-16t^2+160t+0 \\0=-16t^2+160t \\0=-16t(t-10)[/tex]

Using the zero product property, we know that either -16t =0 or t-10=0. When -16t=0 is at t=0, when the rocket is launched. t-10=0 gives us an answer of t=10, so the rocket reaches the ground again at 10 seconds.

[tex]h(t)=-16t^2+v_0 t+h_0[/tex], where v₀ is the starting velocity and h₀ is the initial height. Using the velocity and starting height from our problem we have

[tex]h(t)=-16t^2+160t+0[/tex]. The path of this rocket will be a downward facing parabola, so there will be a maximum. This maximum will be at the vertex of the graph. To find the vertex we start out with [tex]x= \frac{-b}{2a}[/tex], which in our case is [tex]x= \frac{-160}{2(-16)}= \frac{-160}{-32} = 5[/tex]. It will take 5 seconds for the rocket to reach its maximum height. Plugging this back into our formula gives us

[tex]h(5)=-16(5^2)+160(5)+0 \\=-16(25)+800 \\=-400+800=400[/tex]

The rocket's maximum height is 400 feet.

We set our formula equal to zero to find the time it takes to hit the ground, then we factor:

[tex]0=-16t^2+160t+0 \\0=-16t^2+160t \\0=-16t(t-10)[/tex]

Using the zero product property, we know that either -16t =0 or t-10=0. When -16t=0 is at t=0, when the rocket is launched. t-10=0 gives us an answer of t=10, so the rocket reaches the ground again at 10 seconds.