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College Algebra · Chapter 9

Projectile Motion

Throw something upward and gravity writes a quadratic. Its height after t seconds is h(t) = −16t² + v0t + h0 — a parabola opening downward. That means two questions you already know how to answer: the vertex gives the maximum height and when it happens, and the positive zero gives the landing time. Launch it below, then step through four problems worked out in full.

The flight

Set the launch

v0 — how fast it leaves your hand, straight up
h0 — how high it starts
Four launches, worked outpick one, then step through it

Where the −16 comes from — and why the vertex is the top

Nothing here is a new rule. It is the parabola you already know, with t in place of x.

Why the leading coefficient is −16

Near the ground, gravity pulls everything downward at about 32 ft/s². Under a constant acceleration, an object's position picks up a term equal to half the acceleration times the time squared. Start with where it began and how fast it was moving, then subtract what gravity takes back:

building the model h(t) = h0 + v0t − 12(32)t²
     = −16t² + v0t + h0   since ½(32) = 16

So the 16 is not a magic number — it is half of 32. In metric units gravity is about 9.8 m/s², so the same reasoning gives −4.9t². Switch the units button above and watch the coefficient change; nothing else about the algebra does.

The model ignores air resistance, which is why a thrown ball matches it well and a dropped feather does not.

Why the vertex is the maximum height

The leading coefficient is negative, so the parabola opens downward and its vertex is the highest point on the whole curve. The vertex time is the standard t = −b/(2a), with a = −16 and b = v0:

time at the top t = −b2a = −v02(−16) = v032

Notice what is missing from that answer: h0. Starting higher does not change when the object peaks — only how high the peak is. Slide h0 above and watch the time-at-top readout hold still.

Now feed that time back in to get the height itself. This is worth doing once symbolically, because the result is a formula you can check every answer against:

maximum height h(v032) = −16(v032)² + v0(v032) + h0
= −16 · v0²1024 + v0²32 + h0   square the fraction: 32² = 1024
= −v0²64 + 2v0²64 + h0   common denominator 64
= v0²64 + h0

Read that last line out loud: the maximum height is the starting height plus v0²/64. Doubling the launch speed does not double the peak — it quadruples the part gravity has to undo. Drag v0 from 32 to 64 above: the rise goes from 16 ft to 64 ft, four times as much, not twice.

Why the landing time is the positive zero

“It lands” means “its height is 0.” So set the model equal to zero and solve. Multiplying through by −1 first makes the quadratic formula friendlier:

solve h(t) = 0 −16t² + v0t + h0 = 0
16t² − v0t − h0 = 0   multiply both sides by −1
t = v0 ± √v0² + 64h032

Two roots come out, and exactly one of them is usable. Here is why that is guaranteed, not luck. For 16t² − v0t − h0 = 0, the product of the two roots is c/a:

product of the roots t1 · t2 = ca = −h016   which is negative whenever h0 > 0

A negative product means the two roots have opposite signs: one positive, one negative. The negative root is the time the object would have left the ground if it had been thrown from ground level and happened to be passing your hand at t = 0. It is algebraically real but outside the domain of the problem, so you reject it. Switch the figure to Key points to see that rejected root drawn on the graph, marked with an ×.

Domain matters here. The model only describes the object from release until it hits the ground, so the meaningful domain is 0 ≤ t ≤ tland. Any solution outside that interval — negative times, or times after it has already landed — gets discarded, even though it satisfies the equation. Always finish by asking “is this time inside the flight?”

Also notice the discriminant v0² + 64h0 is a sum of two things that cannot be negative, so it is never negative. There is always a real landing time. That is the algebra agreeing with the obvious: what goes up comes down.

The hang-time shortcut, and when it breaks

If the object starts and ends at the same height — typically h0 = 0 — the parabola is symmetric about the vertex, so the trip down takes exactly as long as the trip up:

only when h0 = 0 tland = 2 · ttop = v016

Launch from a cliff or a rooftop and that shortcut is wrong — the fall is longer than the rise, because the object keeps going past its starting level. Set h0 = 0 above and the ledger will show up-time and down-time matching; raise h0 and watch them split apart.

One last trap: this graph is not the path

The curve on the left is height plotted against time. The object itself goes straight up and straight back down — it never moves sideways. The horizontal axis is seconds, not feet. Switch the figure to Real flight to see the same launch as the object actually experiences it: one vertical line, bunched up near the top where it slows down, spread out near the ground where it is moving fast.