NASA launches a rocket at t = 0 t = 0 seconds. Its height, in meters above sea-level, as a function of time is given by h ( t ) = − 4.9 t 2 + 136 t + 325 , what time does splashdown occur?

Answer 1

h(t) =-4.9t^2+136t +365 = 0

use the quadratic formula

t=-b/2a+(1/2a)sqr(b^2-4ac) with a=-4.9, b=136, c=365

t=136/9.8 +(1/9.8)sqr(136^2 +4(4.9)(365)) ignore the negative square root

use a calculator

= about 30.22 seconds for splashdown time