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 equations = - 2.71% + 40t+6.5, where t is the number of
seconds after the ball was thrown. Complete parts a and b.
sed it.
a. After how many seconds is the ball 16 ft above the moon's surface?
After
seconds the ball will be 16 ft above the moon's surface.

Answer 1

To find the time when the baseball on the moon is 16 feet above the surface, we solve the quadratic equation -2.71t^2 + 40t - 9.5 = 0, using the quadratic formula. We take the positive value of 't' to get the time as time cannot be negative.

Step-by-step explanation:

The student's question is about solving a quadratic equation to determine the time it takes for a baseball thrown on the moon to reach a certain height above the surface. Assuming the given equation is s = -2.71t^2 + 40t + 6.5, where s is the height of the ball in feet and t is the time in seconds, we are looking to find the time when s = 16 feet.

To solve this, we set the equation equal to 16 feet and solve for t:

16 = -2.71t^2 + 40t + 6.5

Now, move all terms to one side to set the equation to zero:

-2.71t^2 + 40t - 9.5 = 0

Using the quadratic formula, t = (-b ± √(b^2 - 4ac)) / (2a), where a = -2.71, b = 40, and c = -9.5, we find the two possible times t when the ball is at a height of 16 feet.

The use of the quadratic formula yields two solutions, however, we take the positive value since time cannot be negative, and we assume that the first positive encounter at 16 feet is when the ball is on the way up.