Can I please get help on this word problem? :)

Question

NASA launches a rocket at  t = 0  seconds. Its height, in meters above sea-level, as a function of time is given by  h ( t ) = − 4.9 t 2 + 259 t + 367 .  Assuming that the rocket will splash down into the ocean, at what time does splashdown occur?  How high above sea-level does the rocket get at its peak?

Answer 1

h(t)=-4.9t²+259t+367.

h(t)=0 at splashdown: t=(-259±√(259²+4×4.9×367))/(-9.8)=54.24 seconds. Ignore negative solution for t.

Using calculus dh/dt=-9.8t+259=0 at maximum height. Therefore t=259/9.8=26.43 seconds.

h(26.43)=3789.5 metres.

So it reaches a height of 3789.5m after about 26.43 seconds. Splashdown occurs at about 54.24 seconds.