Question

Question

During a softball game, a batter hits a ball upward from an initial height of 3 feet. The height, in feet, of the softball is given by s(t)=−16t2+80t+3
, where t is time in seconds and t≥0
. Which is closest to the time when the softball will be 50 feet above the ground?(1 point)
Responses

Answer 1

To find the time when the softball will be 50 feet above the ground, we need to set the height function equal to 50 and solve for t:

s(t) = -16t^2 + 80t + 3

Setting this equal to 50, we get:

-16t^2 + 80t + 3 = 50

Rearranging the equation:

-16t^2 + 80t - 47 = 0

To solve this quadratic equation, we can use the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / 2a

In this case, a = -16, b = 80, and c = -47. Plugging these values into the quadratic formula, we get:

t = (-80 ± √(80^2 - 4(-16)(-47))) / 2(-16)

Simplifying the equation:

t = (-80 ± √(6400 - 3016)) / (-32)

t = (-80 ± √(3384)) / (-32)

Since the problem specifies that t ≥ 0, we can ignore the negative value from the ± symbol. So, we have:

t = (-80 + √3384) / (-32)

Calculating the square root:

t = (-80 + √(4 * 846)) / (-32)

t = (-80 + 2√846) / (-32)

Approximating the value of square root of 846:

t ≈ (-80 + 2 * 29.09) / (-32)

t ≈ (-80 + 58.18) / (-32)

t ≈ -22.18 / (-32)

t ≈ 0.693 seconds

Therefore, the time when the softball will be 50 feet above the ground is closest to 0.693 seconds.