Question

Rewrite the following, 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!). 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

Corrected and rewritten:

If I kick a soccer ball straight up with initial speed 32 ft/s, the height y(t) (in feet) after t seconds is
y(t) = -16t^2 + 32t + c,
where t is time in seconds, -16t^2 comes from gravity (in feet units), 32t is the initial upward velocity term, and c is the constant initial height (the value of y at t = 0).

In the quadratic form ax^2 + bx + c:
- a = -16 (gravity term),
- b = 32 (initial velocity),
- c = initial height (the constant).

If the ball is kicked from the ground, c = 0, so y(t) = -16t^2 + 32t. Setting y = 0 to find when it lands:
-16t^2 + 32t = 0 → t( -16t + 32 ) = 0 → t = 0 s (launch) or t = 2 s (landing).
So it returns to the ground after 2 seconds.

If you start from a nonzero initial height, e.g. c = 3 ft, then y(t) = -16t^2 + 32t + 3 and you solve that quadratic (using the quadratic formula) to find the landing time (≈ 2.09 s for c = 3 ft).