A rocket is launched from the top of an 80-foot cliff with an initial
velocity of 88 feet per second. The height of the rocket t seconds
after launch is given by the equation h=-16ť +88t + 80. How high
will the rocket be after 2 seconds?

Step-by-step explanation:

In order to solve this we need to simply sub in a 2 for t.  That looks like this:

which will give us the height at 2 seconds.

Solving, f(2) = 192 feet

192 feet.

Step-by-step explanation:

h)t) = -16t^2 + 88t + 80 therefore:

h(2) = - 16(2)^2 + 88*2 + 80

= -64 + 176 + 80

= 112 + 80

= 192 feet.

step-by-step explanation:

step-by-step explanation:

use pothagarean therom

a^2+b^2=c^2 so first ac

acm is a right triangle because c= 90°

and am (the hypotenuse or c) =16+9 =25

but there is not enough info we need another angle or side length