I need help solving the 2nd question

Question

NASA launches a rocket at 𝑡=0t=0 seconds. Its height, in meters above sea-level, as a function of time is given by ℎ(𝑡)=−4.9𝑡2+280𝑡+220h(t)=−4.9t2+280t+220. 

Assuming that the rocket will splash down into the ocean, at what time does splashdown occur? 

The rocket splashes down after  seconds. 

How high above sea-level does the rocket get at its peak? 

The rocket peaks at  meters above sea-level

Answer 1

h(t) = -4.9t^2 +280t +220

peak height is when v(t)=h'(t) = 0

-9.8t +280= 0

t = 280/9.8 = about 28.57 seconds at max height

max height = h(140/4.9) = -4.9(140/4.9)^2 + 280(140/4.9) +220

= -140^2/4.9 + 2(140^2)/4.9 + 220

= 140^2/4.9 + 220

= 4,220 meters

splash down is a little more than twice 28.57 seconds > 57.14 s

if it had started at ground zero, it would have be exactly twice 28.57s

-4.9t^2 +280t +220 = 0

use the quadratic formula

t = 280/9.8 + (1/9.8)sqr(280^2 +4(4.9)(220))

= (280 + 298.60)/9.8

= about 57.91 seconds

the height function is a downward opening parabola. top of the parabola is the vertex = max point which has a tangent line with slope = 0 = v(t)=h'(t)