Vertical Motion: Time, Height & Acceleration

What's Really Going On Here

The three questions look different, but they are one idea viewed three ways. The apple moves under a constant acceleration, and everything it does is dictated by that single number. The most instructive part is the last question, because most students instinctively give the wrong answer to it.

• Reading hidden information from the story. "Maximum height" means velocity = 0. "Returns to her hand" means displacement = 0. The problem never states either number, but both are given.
• Choosing and keeping a sign convention. Once "up is positive" is chosen, gravity is −9.80 for the whole trip, rising or falling.
• Using symmetry to save work. Rising and falling are mirror images, so half the trip tells you the whole trip.
• Separating velocity from acceleration. The velocity at the top is zero, yet the acceleration is not. This distinction runs through all of mechanics.
• Solving one problem three ways. Kinematics, energy conservation, and the parabola h(t) must all agree, which is a strong check on your work.

Picture This: Height Against Time

Before computing anything, look at what the apple's motion looks like when plotted. The graph below shows height above the hand as a function of time. It is the path of the apple unrolled in time, not a sideways trajectory.

The graph is a parabola, symmetric about the peak. The slope of the curve is the velocity: positive on the way up, zero at the top, negative on the way down. The way the slope itself keeps decreasing is the constant downward acceleration.

Solution: Method 1 — Kinematics with Symmetry

Free flight is the textbook case of constant acceleration, so we use the standard kinematic equations. We take g = 9.80 m/s², as is standard in most courses.

Step 1 — Choose a sign convention and list what you know

Take upward as positive. Gravity pulls down, so the acceleration is negative for the entire flight. Two pieces of information are hidden in the wording, so write them down explicitly.

Step 2 — Find the time to reach the top

The velocity changes at a steady rate, so use v = v₀ + at. At the peak v = 0:

In words: gravity removes 9.80 m/s of upward speed every second, so it takes about a third of a second to remove all 3.25 m/s.

Step 3 — Get the full flight time

The fall mirrors the rise: the apple passes through each height with the same speed going down as going up. So it must return to the hand moving at v = −3.25 m/s. Apply the same velocity equation to the whole trip:

This is exactly twice the rise time from Step 2, 2 × 0.3316 = 0.6633 s, which confirms the symmetry argument.

Step 4 — Find the maximum height

Here time is not needed, so use the time-free equation v² = v₀² + 2aΔy with v = 0:

The height is measured above her hand, which is exactly what the question asks for and exactly what the equation gives, since h is measured from the launch point.

Step 5 — The acceleration at the top

Once the apple leaves her hand, the only force acting on it is gravity (ignoring air resistance). The acceleration is therefore the same at every moment of the flight: a = −9.80 m/s², which is 9.80 m/s² directed downward. The velocity is momentarily zero at the peak, but the rate of change of velocity is not.

Solution: Method 2 — Energy Conservation

This approach comes from a different physical principle. Instead of tracking velocity second by second, it compares the apple at two instants: leaving the hand, and sitting at the top.

Step 1 — Height from energy

At launch the apple has only kinetic energy. At the top it has only gravitational potential energy (speed is zero). With no air resistance, energy is conserved:

Step 2 — Time from the average speed

During the rise the speed drops steadily from 3.25 m/s to 0, so the average speed is the midpoint, 1.625 m/s. Distance equals average speed times time:

Step 3 — Acceleration from the force

Newton's second law says the acceleration is determined by the net force. The only force on the apple in flight is its weight mg, so a = F/m = g = 9.80 m/s² downward, at the peak just as everywhere else.

All three answers match Method 1, but they came from different physical reasoning. That agreement is itself strong evidence they are right.

Solution: Method 3 — The Position Parabola

Method 1 solved for one thing at a time. Here we write the height as one function of time and read off everything from it. This is the graph from earlier, now in algebraic form.

Step 1 — Write the position function

With the hand as the origin, the general constant-acceleration formula h = v₀t + ½at² becomes:

Step 2 — Flight time: where does the parabola hit zero?

The two roots are the two moments the apple is at hand level: the instant of release and the instant of return. The flight time is the second root.

Step 3 — Maximum height: the vertex

For at² + bt + c, the vertex sits at t = −b/(2a). Here that is t = −3.25 ÷ (2 × −4.90) ≈ 0.3316 s, which is exactly halfway between the two roots. Substituting:

Step 4 — Acceleration: read the t² coefficient

The coefficient of t² is ½a. Since ½a = −4.90, we get a = −9.80 m/s². This coefficient is constant, which is why the acceleration is the same at the peak as anywhere else. Differentiating twice gives h″(t) = −9.80, the same result for students who have seen calculus.

Verification

Does This Seem Reasonable?

Half a meter and two-thirds of a second is about what you see when you flip an apple gently from your palm. That is a soft toss, and 3.25 m/s is only about 7 mph. The table shows the whole trip at quarter-flight intervals. Watch the acceleration column.

Velocity changes by the same amount in each interval, about 1.63 m/s per quarter-flight, and the height is symmetric around the peak. The table also shows why the third answer is what it is: the acceleration column never changes.

What Trips People Up

How to Spot This Problem Type

These phrases signal free-flight kinematics:

• "Tosses straight up" / "thrown vertically" / "dropped": constant acceleration of 9.80 m/s², and only gravity acts.
• "Maximum height" / "highest point" / "at the top": the velocity is zero there. This is the hidden given.
• "Returns to her hand" / "back at the starting point": the displacement is zero, and the final speed equals the initial speed.
• "Acceleration at the peak": a conceptual trap, and the answer is always g.

The same structure appears in algebra courses as "a ball's height is given by h = −4.9t² + v₀t," and in physics courses as "constant acceleration." It is the same quadratic with different labels, and the vertex of that quadratic is the maximum height. Recognizing that connection makes a physics problem feel like algebra you already know.

The Pattern Behind This

Solving with letters instead of numbers gives two formulas that work for any vertical toss that returns to its starting height:

Look at how the numbers sit inside these formulas, because that is where the insight lies. Time is proportional to v₀: double the launch speed and the flight lasts twice as long. Height is proportional to v₀²: double the launch speed and the apple rises four times as high. Strengthening a toss pays off more in height than in time. It also explains why a ball thrown at 10 m/s goes dramatically higher than one thrown at 5 m/s.

Applying it to our case: T = 6.50/9.80 = 0.663 s, H = 10.5625/19.6 = 0.539 m, and the third formula agrees, 9.80 × 0.4399 ÷ 8 = 0.539 m. These formulas hold only when the object lands at the height it started from and air resistance is negligible. For a toss from a rooftop (see the What-If problems), you must return to the quadratic.

Beyond the Textbook

• Measuring vertical jumps. Sports scientists use contact mats that record only how long an athlete is airborne. The formula H = gT²/8 converts that single number into jump height. A half-second hang time means about 31 cm.
• Fountains and water jets. Designers choose the nozzle speed using H = v₀²/(2g). A fountain that must reach 5 m needs the water to leave at about 9.9 m/s.
• Space and planetary science. The same equations, with a different g, predict how high a rover's dropped tool or a thrown stone will go on the Moon or Mars. The What-If section below tries it.

A Brief History

For nearly two thousand years, the dominant view followed Aristotle: heavier objects fall faster, and a thrown object needs a continual push to keep moving. Around 1600 Galileo challenged this with careful experiments using inclined planes and pendulums, finding that falling bodies gain speed at a constant rate regardless of weight. His result, that distance fallen grows with the square of time, is the origin of the ½gt² term in our parabola. Newton later explained why the acceleration is constant: a steady force acts on the object throughout its flight. The point about the peak, that zero velocity does not mean zero acceleration, was one of the conceptual shifts of that revolution.

Where This Leads

Two natural extensions follow. First, toss the apple at an angle instead of straight up: the vertical part behaves exactly as in this problem, while the horizontal part moves at constant speed, which gives the parabolic paths of projectile motion. Second, include air resistance, so the acceleration depends on velocity and the rise and fall are no longer symmetric. That problem is typically handled with differential equations. A good first step toward it: at the peak, what direction does air drag point on the way up versus the way down?

Extend Your Thinking

Four variations, in increasing order of difficulty. Try each before you open the solution. Use g = 9.80 m/s² unless a problem says otherwise.