A rocket is launched from atop a 65-foot cliff with an initial velocity of 113 ft/s. The height of the rocket above the ground at time t is given by -16t^2+141t+115 When will the rocket hit the ground after it is launched? Round to the nearest tenth of a second.

Question

campbell_st · Answer

it will hit the ground when h(t) = 0... or it has zero height.. 

so you need to solve 

$$-16t^2 + 141t + 115 = 0$$

the general quadratic formula ma come in handy

anonymous · Answer

Which is?

anonymous · Answer

Im just very confused on the whole section. @campbell_st

campbell_st · Answer

$$t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

you have a = -16, b = 141 and c = 115

substitute the values and calculate t. 

only use the positive answer.

anonymous · Answer

thanks :)