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
put in standard form.
16t^2 -88t+72=0
2t^2-11t+9=0
(2t -9)(t-1)=0
t= 4.5, 1
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.