Algebra please help

Question

A rocket is fired with an initial velocity v of 120 meters per second. The quadratic function S(t) = -5t^2 + 120t can be used to find the height s of the rocket, in meters, at any time t in seconds. Find the height of the rocket 8 seconds after it takes off. During the course of its flight, after how many seconds will the rocket be at a height of 670 meters?

Answer 1

The information of the first sentence is actually not needed to calculate the height.

S(t) is also what we call a position function and it can be used to find the height 's' for any arbitrary time 't'.

Therefore, if t=8 seconds, then S(8) = -5(8)² + 120(8) = -320 + 960 = 640 meters.

To answer the second part, we want to solve for 't'. Therefore, if s = 670 meters, then 670 = -5t² + 120t. Subtract the 670 from both sides and divide both sides by -1 for simplicity. Then, 5t² - 120t +670 = 0. Notice that each term is a multiple of 5. Divide both sides by 5, then t² - 24t + 134 = 0. Now we can either factor this quadratic polynomial or use the quadratic equation. Notice that the multiples of 134 (1, 67, 134) do not add up to 24, so we should use the quadratic equation.

Therefore, t = (-(-24) +- ((-24)² - 4(1)(134))^1/2)/2(1)

After some simplification, t = (24 +- (40)^1/2)/2

⇒ t = 12 +- (10)^1/2

⇒ t = 15.16 seconds OR t = 8.84 seconds after the launch of the rocket.

Answer 2

To solve this, first to solve for the height of the rocket 8 seconds after takeoff, you plug 8 into the position equation. Then S(8)= -5*(8^2)+120*8=-320+960= 640 meters. To find after how many seconds the rocket will be at a height of 670 meters, we set S(t)=-5*t^2+120*t= 670, and solve for t. We have the equation 670=-5t^2 +120t, we subtract 670 from both sides then divide through by -5 to get the equation t^2-24t+134=0. We can solve this quadratic in numerous ways, the simplest of which is plugging it into the quadratic formula to find that t= 12±√(576-536)/2= 12±√10 seconds.