a rocket is fired upward with an initial velocity

Question

A rocket is fired upward with an initial velocity v of 80 meters per second. The quadratic function Upper S left parenthesis t right parenthesis equals negative 5 t squared plus 80t S(t)=−5t^2+80t 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 77 seconds after it takes off. During the course of its? flight, after how many seconds will the rocket be at a height of 245 ?meters?

Answer 1

If the rocket is launched from a height of zero,

 

S(t) = -5t² + 80t

 

S(77) = -5(77²) + 80(77) = -23,485 meters 

 

The rocket is underground!  Are you sure the 77 is not a typo?  I suspect you meant 7 seconds.

 

S(7) = -5(7²) + 80(7) = 315 meters

 

For the second part of the problem, set S(t) = 245 and solve for t.

 

245 = -5t² + 80t

 

-5t² + 80t - 245 = 0

 

t² - 16t + 49 = 0

 

t = (1/2) {-(-16) ± √[(-16)² - 4(1)(49)]}

 

t = (1/2)(16 ± √60) = (1/2)(16 ± 7.746)

 

The positive square root gives t = 11.873 seconds.

 

The negative square root gives t = 4.127 seconds.

 

The lower t is the time it takes to reach 245 meters on the way up, and the higher t is the time it takes to reach 245 meters on the way down.