The height h in feet of a baseball after t seconds is given by h = -16t2 + 88t + 3. At what values of t is the height of the baseball 75 feet?

75=-16t^2+88t+3

put in standard form.

16t^2 -88t+72=0

2t^2-11t+9=0
(2t -9)(t-1)=0

t= 4.5, 1

To find the values of t when the height of the baseball is 75 feet, we need to solve the equation -16t^2 + 88t + 3 = 75.

First, we subtract 75 from both sides of the equation to get -16t^2 + 88t + 3 - 75 = 0, which simplifies to -16t^2 + 88t - 72 = 0.

Now, we have a quadratic equation in the form of at^2 + bt + c = 0, where a = -16, b = 88, and c = -72.

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

t = (-b ± √(b^2 - 4ac)) / (2a)

Plugging in the values, we have:

t = (-88 ± √(88^2 - 4(-16)(-72))) / (2(-16))

Now we simplify further.

t = (-88 ± √(7744 - 4608)) / (-32)

t = (-88 ± √3136) / (-32)

t = (-88 ± 56) / (-32)

This gives two solutions:

t1 = (-88 + 56) / (-32) = -32 / (-32) = 1

t2 = (-88 - 56) / (-32) = -144 / (-32) = 4.5

Therefore, the height of the baseball is 75 feet at t = 1 second and t = 4.5 seconds.