Question

For the following, rewrite 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!). You can 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

Write the height as a quadratic in the form ax^2 + bx + c (here x = t):

y(t) = -16t^2 + 32t + c.

- a = -16 (ft/s^2) — the acceleration from gravity (determines the downward curvature).
- b = 32 (ft/s) — the initial upward velocity.
- c = initial height (ft) — the constant. For your scenario, kicked from the ground, c = 0, so

y(t) = -16t^2 + 32t.

Setting y = 0 gives -16t^2 + 32t = 0 → t( -16t + 32 ) = 0 → t = 0 s (launch) or t = 2 s (landing). The ball returns to the ground after 2 seconds.

(If instead the ball started on a platform, pick c > 0; e.g. c = 5 ft gives y = -16t^2 + 32t + 5, and you would solve that quadratic for landing time.)