A model rocket is launched from the top of a building. The height(in meters) of the rocket above the ground is given by h(t)=-6t^(2)+24t+14, where t is the time (in seconds) since the rocket was launched. What is the rocket's maximum height?

Step-by-step explanation:

I don't know if you are a calculus student, but since calculus is the easiest way to solve this, that's what I used.  The position equation is

and the velocity function, the first derivative, is

From Physics, you should know that the velocity of an object is 0 at its highest point.  That means that if we sub in a 0 for the velocity and solve for t, we find the time at which the object is at its highest point.  That is:

and

By the Zero Product Property,

-12 definitely does not equal 0, so that means that

t - 2 = 0 and

t = 2 seconds.

So at 2 seconds, the object is at its highest point.  Go back to the position function now and sub in a 2 for t to find the height of the object at 2 seconds.

so

s(2) = 38 meters

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2

the correct answer is 5 6 and 7