The quadratic function h(t)-16t^2+112t+128 shows the ball's height about the ground, h(t), in feet, t, seconds after it was thrown

Question

A person standing close to the edge on top of a 128-foot building throws a ball vertically upward. The quadratic function h(t)-16t^2+112t+128 shows the ball's height about the ground, h(t), in feet, t, seconds after it was thrown

What is the maximum height of the ball?

How many seconds does it take until the ball hits the ground?

Answer 1

To solve for the vertex (maximum) of a parabola we can use the formula

1. t (max) = -b/2*a, when the form of the parabola is h(t) = a*t^2 + b*t + c
2. Applying this formula,
3. a = -16
4. b = 112
5. -b/2*a = -(112)/((2)*(-16)) = 112/32 = 3.5
6. Now, we know that the maximum height is reached at t = 3.5. We need to plug this value in to determine the maximum height:
7. h(3.5) = -16*(3.5)^2+112*(3.5)+128 = -196 + 392 + 128 = 324 feet
8. To understand how long it takes the ball to hit the ground we use our knowledge that the ground is represented by h(t) = 0. We plug this into our formula to solve for t:
9. 0 = -16t^2 + 112t + 128 = -16*(t^2 - 7t - 8) = -16*(t - 8)*(t + 1)
10. So now we see the roots of the function are t = 8 and t = -1
11. However, time cannot be negative, which results in the conclusion that the ball hits the ground at t = 8