Question

Rewrite the following below, and state what part of the problem the constant is (remember the format is ax^2 + bx + c. where c is the initial value, which is the constant!). add a value to represent your constant for the scenario.

I kick a soccer ball straight up from the ground with an initial speed of 32 feet per second, and I want to know when it will land. The height y (in feet) after t seconds is
y + 32t, and solving y = 0 shows the ball returs to the ground after after seconds. In this function t is the time in seconds, the number 16 comes from gravity (i
would use - 16t^2 when measuring in feet), and the 32 is the ball's starting upward speed.

Answer 1

Corrected model and solution:

The height y (in feet) after t seconds is
y(t) = -16t^2 + 32t + c,
where a = -16, b = 32, and c is the constant (the initial height, i.e. the initial value).

- a = -16 comes from gravity when using feet (−16t^2).
- b = 32 is the initial upward speed (32 ft/s).
- c is the initial height above the ground (the constant).

For the scenario “kicked from the ground,” take c = 0, so
y(t) = -16t^2 + 32t.
Solve y = 0: -16t^2 + 32t = 0 ⇒ t(−16t + 32) = 0 ⇒ t = 0 s or t = 2 s.
The ball returns to the ground after 2 seconds.

(If you wanted a nonzero constant, e.g. c = 1 ft initial height, use y(t) = -16t^2 + 32t + 1 and solve the quadratic; that gives the positive landing time ≈ 2.03 s.)