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 equals negative 2.7 t squared plus 40 t plus 6.5s=−2.7t2+40t+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 18 ft above the​ moon's surface?

Answer 1

Answer- 0.29s

Solution-

The given astronaut's height and the initial velocity of the ball are extra data, they are not necessary for calculation.

The initial velocity is irrelevant because our equation already compensates for it as it gives the height of the ball, and we see that the equation also has 6.5 at the end, which is the astronaut's height, in feet. So it's already added.

So Height or Displacement = S = -2.7t² + 40t + 6.5

⇒ 18 = -2.7t² + 40t + 6.5

⇒ 2.7t² - 40t + 11.5 = 0

Now solving the equation using the quadratic formula,

$t=\frac{-b\pm \sqrt{b^{2}-4ac } }{2a}$

⇒ $t=\frac{-(-40)\pm \sqrt{(-40)^{2}-4(2.7)(11.5) } }{2(2.7)} = 0.29 \ or 14.52$

It passes 12ft on the upward path after 0.29s, and falls downward to a position of 12ft after 14.52s.