If a metal ball is thrown downward with an initial velocity of 22 feet per second (15mph)from a 100 foot water tower, it's height h in feet above the ground after t seconds is modeled by h(t)=-16t^2-22t+100
Determine when the height of the ball is 62 feet.
h = 100 - 22 t - 16 t^2
62 = 100 - 22 t - 16 t^2
16 t^2 + 22 t - 38 = 0
8 t^2 + 11 t - 19 = 0
(8 t + 19) (t - 1) = 0
t = 1 second (forget the negative time)
To determine when the height of the ball is 62 feet, we need to solve the equation h(t) = 62.
The given equation is h(t) = -16t^2 - 22t + 100.
To solve the equation, we substitute h(t) with 62:
62 = -16t^2 - 22t + 100
Rearranging the equation, we get:
16t^2 + 22t - 38 = 0
Now we have a quadratic equation in the form of at^2 + bt + c = 0, where a = 16, b = 22, and c = -38.
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values, we get:
t = (-22 ± √(22^2 - 4(16)(-38))) / (2(16))
Simplifying:
t = (-22 ± √(484 + 2432)) / 32
t = (-22 ± √2916) / 32
t = (-22 ± 54) / 32
We have two solutions:
1. t = (-22 + 54) / 32 = 32 / 32 = 1
2. t = (-22 - 54) / 32 = -76 / 32 = -2.375
Therefore, the height of the ball is 62 feet at 1 second and -2.375 seconds. However, since time cannot be negative in this context, we discard the negative solution.
So, the height of the ball is 62 feet after 1 second.