solve using quadratic equation

Question

An astronaut on the moon throws a baseball upward. The astronaut is 6​ ft, 6 in.​ tall, and the initial velocity of the ball is 30ft per sec. The height s of the ball in feet is given by the equation s=-2.7t^2+30t+6.5 ​, where t is the number of seconds after the ball was thrown. Complete parts a and b. a. After how many seconds is the ball 12ft above the​ moon's surface? After nothing seconds the ball will be 12 ft above the​ moon's surface. ​(Round to the nearest hundredth as needed. Use a comma to separate answers as​ needed.)

Answer 1

h(t) =-2.7t^2 +30t + 6.5

12 = -2.7t^2 + 30t + 6.5

2.7t^2 -30t + 5.5 = 0

use the quadratic formula

t = 30/5.4 + or - (1/5.4)sqr(900- 4(2.7)(5.5))

t = 0.1865 seconds or 10.925 seconds

Answer 2

Here, we are asked to find after how many seconds (t) is the ball's height (s) 12 feet above the moon's surface.

Using the equation:

s(t) = -2.7t^2 + 30t + 6.5

When s=12, rearrange the equation to make it equal to 0:

12 = -2.7t^2 + 30t + 6.5

0 = -2.7^2 + 30t - 5.5

Using the quadratic formula, plug in the values a = -2.7, b = 30, c = -5.5

x = [ -b +/- sqrt(b^2 - 4ac) ] / 2a

x = [ -30 +/- sqrt (30^2 - 4*-2.7*-5.5) ] / 2*-2.7

solve for the 2 values of x:

x = [ -30 +/- sqrt(840.6) ] / -5.4

x = [ -30 +/- 28.99 ] / -5.4

x = 0.19, 10.92

The reason there are two solutions is because when you throw a ball in the air it goes up first, then down. When you throw it up it will reach 12 feet on the way up after a short time, then again as the ball is on the way down after a longer time.