help i need an answer asap

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+58t+178h(t)=-4.9t2+58t+178.

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

Rounding to 3 decimal places, the rocket splashes down after _______ seconds.

Rounding to the nearest meter, 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 + 58t + 178

take the derivative and set = 0 to solve for time of max height

h'(t) = v(t) = -9.8t + 58=0

t = 58/9.8 = about 5.92 seconds to reach maximum height

plug back into the h(t) equation to solve for max height

h(58/9.8) = -4.9(58/9.8)^2 + 58(58/9.8) + 178

= about -171.633 + 343.265 + 178

= about 349.633 meters = max height

for splash down time

set h(t)=0 = -4.9t^2 +58t +178

4.9t^2 -58t -178=0

use the quadratic formula

t = 58/9.8 + (1/9.8)sqr(58^2+4(4.9)(178))

= (58 + sqr(6852.8))/9.8

= 14.365 seconds to splash down