Question

During a soft ball game, a batter hits a ball upward from an initial height of 3 feet. The height, in feet, of the soft ball is given by s(t) = -16t^2 + 70t + 3, where t is time in seconds, and t greater than or equal to 0. At what time will the softball be 50 feet above the ground?

Answer 1

To find the time at which the softball will be 50 feet above the ground, we need to solve the equation s(t) = 50.

-16t^2 + 70t + 3 = 50

Rearranging the equation, we get:

-16t^2 + 70t - 47 = 0

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

t = (-b ± sqrt(b^2 - 4ac))/(2a)

For our equation, a = -16, b = 70, and c = -47.

Plugging these values into the quadratic formula, we get:

t = (-70 ± sqrt(70^2 - 4(-16)(-47)))/(2(-16))

Simplifying and calculating the discriminant:

t = (-70 ± sqrt(4900 - 2992))/(-32)
t = (-70 ± sqrt(1908))/(-32)
t = (-70 ± 43.69)/(-32)

Simplifying further, we have two possible values for t:

t = (-70 + 43.69)/(-32) = 1.0703 seconds
t = (-70 - 43.69)/(-32) = 3.1406 seconds

Therefore, the softball will be 50 feet above the ground at approximately 1.0703 seconds and 3.1406 seconds.