Answer:
e. It will take 11 seconds to reach the maximum height of 1,936 feet.
f. It will take 22 seconds to return to the earth.
Step-by-step explanation:
Given:
Initial velocity [tex]v_0[/tex] = 352 ft/sec
Solving for question e.
To find the time required to reach the maximum height we will use the formula,
[tex]h(t) = -16t^2+v_0t+h_0[/tex],
where [tex]v_0[/tex] is the starting velocity
[tex]h_0[/tex] is the initial height.
Using the velocity and starting height from our problem we have,
[tex]h(t) = -16t^2+352t+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{-352}{2(-16)}=\frac{-352}{-32}= 11[/tex]
So, It will take 11 seconds for the rocket to reach its maximum height.
We will find maximum height using the formula by substituting value of t we get,
[tex]h(11)=-16(11^2)+352(11)+0\\h(11) = -16 \times121+ 352 \times 11 = -1936+3872= 1936 \ ft[/tex]
Hence the maximum height will be [tex] 1936 \ ft[/tex]
Now Solving for question f.
To find the time required for rocket to reach earth.
We will set our formula to 0 to find the time.
[tex]0= -16t^2+352t+0\\-16t(t-22)=0[/tex]
Using the zero product property, we know that either -16t = 0 or t - 22 = 0. When -16t = 0 is at t = 0, when the rocket is launched. t - 22 = 0 gives us an answer of t = 22.
So the rocket reaches the Earth again at 22 seconds.
