Question

An astronaut on the moon throws a baseball upward. The astronaut is 6 ft, 6 in. talk and the initial velocity of the ball is 40 ft per sec. The height s of the hall in feet is given by the equation s=-2.7t^2 + 40t+6.5 , where the number of seconds after the ball was thrown. After how many seconds in the ball 22ft above the moons surface?

Answer 1

Explanation:

Using the given equation, we can solve for t when s is equal to 22.

22 = -2.7t^2 + 40t + 6.5

Subtracting 6.5 from both sides, we get:

15.5 = -2.7t^2 + 40t

Adding 2.7t^2 to both sides, we get:

18.2t^2 = 15.5 + 40t

Subtracting 40t from both sides, we get:

18.2t^2 - 40t = 15.5

Dividing both sides by 18.2, we get:

t^2 - 2.2t = 0.855

Factoring, we get:

(t - 1)(t - 0.855) = 0

Setting each factor equal to 0 and solving, we get:

t = 1 and t = 0.855

Therefore, the ball will reach a height of 22ft after 1 second and 0.855 seconds.