algebra 1 that resolution

Question

A rocket is launched at t=0 seconds. Its height, in meters above sea-level, as a function of time is given by h=-4.9t2+346t+290.

At what time does the rocket hit the water? (Round answer to 2 decimal places.)

The rocket splashes down after seconds.

Answer 1

The rocket is launched at t=0 from a height of h = 290 meters. It goes up, then down, until it splashes into sea-level water (h=0).

So, we need to find the value of t when:

0 = -4.9t² + 346t + 290

The quadratic formula is:

t = (-b ± √(b² - 4ac) ) / (2a)

t = (346 ± √(346*346 - 4*(-4.9)*(290) ) / (2*(-4.9))

So, either t = -0.82843 (we reject negative)

or t = 71.4407 sec

Check: put t = 71.4407 into the formula for h and get h = -0.00852

Rounded, t = 71.44 sec

Answer 2

Two ways to work this problem.

The first is to convert to vertex form using completion of the square.

Too much work!

The second is the use the axis of symmetry.

At the axis of symmetry the rocket has reached the highest point and is returning.

t = -346 / 2(-4.9)

t = 35.306

That is only half way. The rocket take the same time to descend to the water

The flight takes 2(35.306) seconds.