Question

You throw a softball straight up into the air with an initial vertical velocity of 40 feet per second. The function represents the height h (in feet) of the softball after t seconds. a) After how many seconds is the height 15 feet?

Answer 1

To find the time when the height of the softball is 15 feet, we can use the quadratic formula to solve the equation derived from the given function. The resulting solutions are approximately 1.25 seconds and 1.75 seconds.

Step-by-step explanation:

To find the time when the height of the softball is 15 feet, we can use the equation h(t) = -16t^2 + 40t, where h(t) is the height in feet and t is the time in seconds.

Substitute 15 for h(t): 15 = -16t^2 + 40t

Rearrange the equation: 16t^2 - 40t + 15 = 0

Use the quadratic formula to solve for t: t = (-b ± sqrt(b^2 - 4ac)) / (2a)

Plug in the values: t = (-(-40) ± sqrt((-40)^2 - 4(16)(15))) / (2(16))

Simplify the equation: t = (40 ± sqrt(1600 - 960)) / 32

Calculate the square root: t = (40 ± sqrt(640)) / 32

Simplify the square root: t = (40 ± 8sqrt(10)) / 32

Simplify further: t = 5/4 ± sqrt(10)/4

So, the height of the softball is 15 feet after approximately 1.25 seconds or 1.75 seconds.