contestada

The altitude of a model rocket launched into the air from a rooftop is given by the quadratic equation A(t) = −16t2 + 32t + 48, where t is the time in seconds since launch, and A is measured in feet. At what time does the rocket land on the ground?

Respuesta :

Answer:

Rocket will land on the ground in 3 seconds.

Step-by-step explanation:

Given:

[tex]A(t) = -16t^2 + 32t + 48[/tex]

where [tex]t[/tex] ⇒ time in seconds since launch.

[tex]A(t)[/tex] ⇒ altitude of the rocket after reaching the ground.

we need to find the time at which rocket will land on the ground.

Solution:

Now we can say that;

altitude of the rocket after reaching the ground will be equal to 0.

So;

[tex]A(t)=0[/tex]

Now Substituting [tex]A(t)=0[/tex] in given expression we get;

[tex]0=-16t^2+32+48[/tex]

Now we take -16 common we get;

[tex]0=-16(t^2-2t-3)[/tex]

Now Dividing both side by -16 we get;

[tex]\frac0{-16}=\frac{-16(t^2-2t-3)}{-16}\\\\0=t^2-2t-3[/tex]

Now factorizing the above equation we get;

[tex]0=t^2-3t+t-3\\\\0=t(t-3)+1(t-3)\\\\0=(t+1)(t-3)[/tex]

Now we will find 2 values of t by substituting each separately.

[tex]t+1=0 \ \ \ Or \ \ \ \ t-3 =0\\\\t =-1 \ \ \ \ \ \ Or \ \ \ \ \ \ t=3[/tex]

Now we get 2 values of t one positive and one negative.

Now we know that time cannot e negative hence we will discard it and consider positive value of t.

Hence Rocket will land on the ground in 3 seconds.