a) 10 sec
b) 256 feet
Step-by-step explanation:
a)
The height of the rocket above the ground at time t is described by the second-order equation
[tex]h(t)=160t-16t^2[/tex] (1)
where:
t is the time, in seconds
h(t) is the height, in feet
[tex]160[/tex] represents the initial vertical velocity of the rocket
[tex]-16[/tex] represents the acceleration due to gravity (downward)
The rocket will return to the ground when the height is zero, so when
[tex]h(t)=0[/tex]
Substituting into eq(1) and solving for t, we find:
[tex]0=160t-16t^2\\0=16t(10-t)=0[/tex]
which has two solutions:
t = 0 (initial instant, so when the rocket starts its motion)
t = 10 s --> this is the time at which the rocket returns to the ground
b)
The height of the rocket is given by
[tex]h(t)=160t-16t^2[/tex]
t is the time, in seconds
h(t) is the height, in feet
Here we want to find the height of the rocket after 2 seconds: to do that, we just need to substitute
t = 2 sec
into the equation of the height.
By doing so, we find:
[tex]h(2)=160\cdot 2 -16(2)^2=256 ft[/tex]
Therefore, the height of the rocket after 2 seconds is 256 feet.
