After how many seconds is the ball 20 ft above the​ moon's surface?

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 40 ft per sec. The height s of the ball in feet is given by the equation s=-2.7t^2+40t+6.5​, where t is the number of seconds after the ball was thrown. Complete parts a and b.

Answer 1

s = -2.7t² + 40t + 6.5​

20 = -2.7t² + 40t + 6.5​

-2.7t² + 40t - 13.5​ = 0

Using the quadratic formula, for ax² + bx + c = 0 then x = [-b ± √(b² - 4ac)]/(2a), then:

t = [-40 ± √((40)² - 4(-2.7)(-13.5))]/(2(-2.7))

t = [-40 ± √1452.2]/(-5.4)

t = (-40 ± 38.134)/(-5.4)

t₁ = (-40 + 38.134)/(-5.4) = 0.346

t₂ = (-40 - 38.134)/(-5.4) = 14.469

So after 0.346 seconds the ball goes above 20 feet and then after 14.469 seconds the ball falls back below 20 feet.