A rocket is launched from atop a 74-foot cliff with an initial velocity of 113 ft/s. The height of the rocket above the ground at time t is given by h=16t^2+113t+74. When will the rocket hit the ground after it is launched? Round to the nearest tenth of a second.

9514 1404 393

  7.7 seconds

Step-by-step explanation:

The solution is given by the quadratic formula.

For ax² +bx +c = 0, the roots are ...

  x = (-b±√(b²-4ac))/(2a)

We want the positive root, so ...

  x = (-113 -√(113² -4(-16)(74)))/(2(-16)) = (113 +√17505)/32 ≈ 7.6658

The rocket will hit the ground about 7.7 seconds after it is launched.

answer: haha (: i’ll go see what it is

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