Don't worry. It's explained below.

    The quadratic is a ideal model for projection -- shooting, propelling, dropping stuff at a certain speed, from a certain height.

    The formula s(t) = (a/2)t² + v₀t + s₀

may be used to represent
• s(t) -- the vertical displacement (distance from ground level) over time the time period t,
• where a is the force of gravity,
  Note: Gravity is almost always negative for downward.
• v₀, meaning v(0), the initial velocity, the velocity at time 0, and
• s₀, meaning s(0), the initial displacement, the displacement or height at time 0, the starting time.  
• Wind resistance and the initial trajectory of the object may not be coded in this equation.

    Don't worry. This reminds you.

    On this page, we are not using x and y as the variables. We are using t and s.

    The s stands for height or displacement or DISTANCE ABOVE (OR BELOW) GROUND LEVEL.

    The t stands for TIME.

    Restated, this equation is a model for how a thing moves relative to the ground over a period of time.

    Restated, the horizonal scale measures the TIME and the vertical scale measures the DISTANCE ABOVE OR BELOW GROUND.

    The equation "s(t) = 2" says "The height above ground is always 2."

    The equation "s(t) = 2t" says "The height above ground is twice the time." At starting time, time 0, the height is 0 -- the object is on the ground. After 1 second, the height is 2. After 2 seconds, the height is 4. After 10 seconds, the height is 20.

    The equation "s(t) = s₀ + v₀t + (a/2)t²" says "The height above ground is the sum of the following 3 terms (features): the starting height, the starting velocity times the time, and half the acceleration times the square of the time.

    Look at the constant terms.

s(t) = -16t² + 0t + 60

Starts 60 feet above the ground.

s(t) = -9.1t² + 20t + 0

Starts at ground level.

s(t) = -4t² + 12t + 0

Starts at ground level.

s(t) = -16t² + 40t - 10

Starts 10 feet below ground.

s(t) = -16(t + 1)(t - 3), s(t) = -16t² + 32t + 48

Starts 48 feet above ground.

    Look at the coefficient of the linear term.

s(t) = -16t² + 0t + 60

Starting speed is 0 ft./sec. - it is dropped.

s(t) = -9.1t² + 20t + 0

Starting speed is 20 ft./sec, upward.

s(t) = -4t² + 12t + 0

Starting speed is 12 units/sec., upward.

s(t) = -16t² + 40t - 10

Strating speed is 40 units/sec., upward.

s(t) = -16t² - 40t - 10

Strating speed is 40 units/sec., downward.

s(t) = -16(t + 1)(t - 3), s(t) = -16t² + 32t + 48

Starting speed is 32 ft./sec.

    Look at the coefficient of the quadratic term.

    The power on the variable causes the trajectory, path, of the thing to be a parabola, a quadratic function. It makes the path a U-shaped curve.

    In every equation, the coefficient of the quadratic term is negative. Gravity is bringing the object to the ground even if it is shot upward. The U-shaped curve opens down. If there existed a case in which the more massive object were up, instead of down as on earth, the gravity would be positive and things would not fall but rise and the coefficient of the quadratic term would be positive.

s(t) = -16t² + 0t + 60

This is the gravity of earth, 32 ft./sec.².

s(t) = -9.1t² + 20t + 0

This is not the gravity of earth. Here it's 18.2 ft./sec.².

s(t) = -4t² + 12t + 0

This is not the gravity of earth. Here it's 8 ft./sec.².

s(t) = -16t² + 40t - 10

This is the gravity of earth, 32 ft./sec.².

s(t) = -16(t + 1)(t - 3), s(t) = -16t² + 32t + 48

This is the gravity of earth, 32 ft./sec.².

Dropped, not shot or thrown upward, so no additional height

s(t) = -16t² + 0t + 60

    An object is dropped from 60 feet above the ground under the force of gravity (32 ft./sec.² toward the ground).

    Notice that the path is parabolic even though not all of the parabola describes the situation. The event begins at time 0.

    The graph has been drawn to indicate the object does not travel below ground once returning to earth.

    Object starts at ground level and gravity is -18.2 m/s².

    It is shot upward at 20 m/s.

s(t) = -9.1t² + 20t + 0

    Notice that the path is parabolic even though not all of a parabola describes the situation. The event begins at time 0 and, here it is assumed that, the motion ends once the object hits the ground.

    Starts on the ground. Hits the ground 3 seconds later.

s(t) = -4t² + 12t + 0

    On some planet with gravity lighter than earth's, an object is shot upward at 12 ft. per second, from ground level.

    Examine this case and ALL OTHER CASES OF PARABOLIC PATHS OF FLIGHT. The maximum height is achieved midway between the at-ground-level times. This is always the case. See the diagram at the top of the page.

    The first ground level time is 0. The time it next hits ground is 3 seconds. The total time is 3 seconds. Half way or midway between these times when the height is 0 is 1.5 seconds. The object reaches its maximum height at 1.5 seconds.

    The maximum height is s(1.5), the height at 1.5 seconds. That is:

s(t) = -4t² + 12t + 0 when t is 1.5

s(1.5) = -4(1.5)² + 12(1.5) + 0 = -9+18+0 = 9 ft.

When the object reaches its maximum height of 9 ft. at 1.5 seconds.

 

    Note: This "magically easy to use" trick of finding the maximum height by evaluating the value midway between the two ground points is not unique to the ground points!

    Note: Use the points at a hight of 2 or -1 or 3. They would work as well. The height of 0, the ground height, usually makes the work easier.

s(t) = -16t² + 40t - 10

    The equation tells us the object starts 10 feet below ground, is propelled upward at 40 feet per second, and earth's gravity of 32 feet per second squared is pulling it toward earth.

    The graphs tell more.

   

---

    Ex. 4a. An object is shot into the air from 10 feet below ground and it takes the path described by

s(t) = -16t² + 40t - 10.

    Find the time at which the object finally falls to the ground.

    Using the quadratic equation to solve the equation -16t² + 40t - 10 = 0, the object is at ground level at 2.22 seconds and .28 seconds. The answer is 2.22 seconds.

   

---

    Ex. 4b. An object is shot into the air from 10 feet below ground, surfaces, reaches a maximum height, falls back below ground. It takes the path described by s(t) = -16t² + 40t - 10. Find the time at which the object falls 5 feet below ground.

    Find when the height is -5 feet, 5 feet below ground.

    Solve the equation -16t² + 40t - 10 = -5.

    Using the quadratic equation to solve the equation, the object is 5 feet below ground at 2.37 seconds and .132 seconds. The answer is 2.37 seconds. At .132 seconds the object is still climbing.

   

---

    Ex. 4c. An object is shot into the air from 10 feet below ground and it takes the path described by s(t) = -16t² + 40t - 10.

    Find its height after 2 seconds.

    Find its height after 2 seconds means find s(2).

    Find s(t) = -16t² + 40t - 10 when t is 2.

    Find s(2) = -16(2)² + 40(2) - 10, or 6.

    The object is 6 feet above ground.

s(t) = -16(t + 1)(t - 3), s(t) = -16t² + 32t + 48

    The units feet and seconds were chosen for the example.

    Because the constant term, s₀, is 48, the starting height 48 feet.

    Because the coefficient of the linear term, v(0) or v₀ is 32, the starting velocity is 32 feet per second.

    Because the quadratic coefficient, a/2, is -16, the event occurs on earth or an earth-like place. The object which is lighter than earth will eventually fall to earth.

    Because the roots are t = -1 (1 second before the event happens) and t = 3, the projectile is at ground level at t₃, after 3 seconds.

    Because the midpoint between the roots is 1 second, the maximum height occurs at 1 second.

    Because the value of the function s(t) = -16(t + 1)(t - 3) when t is 1 second is s(1) = -16((1) + 1)((1) - 3) or -16(2)(- 2) or 64, the maximum height is 64 feet.