Arocket is launched into the air. the projectile motion of the rocket can be modeled using h(t) = 112t –16t2, where t is the time since launch in seconds and h(t) is the height of the rocket at time t. when will the rocket be 196 feet in the air?

After 3.5 sec rocket will be at 196 feet in the air.

Step-by-step explanation:

The projectile motion of a rocket can be modeled using h(t) = 112t - 16t²

where t is the time since launch in seconds.

h(t) is the height of the rocket at time t.

we have to calculate when height h is 196 feet.

Now from the given function

196 = 112t - 16t²

-16t² + 112t - 196 = 0

-4t² + 28t - 49 = 0

Or 4t² - 28t + 49 = 0

(2t - 7)² = 0

2t - 7 = 0

t = 7/2 = 3.5 sec

look a the sides they are perpendicular

for this case we have by definition, if two lines are perpendicular, then the product of their slopes is -1.

then, given the following line:

so,

we are looking for m_ {2}:

then, the line is given by:

we find "b" replacing the given point:

finally, the equation is:

option c