Vertical acceleration is the rate at which an object’s vertical velocity changes over time. In most physics problems, it equals 9.8 m/s² downward (the acceleration due to gravity), but when other forces act vertically, you need a different approach. The method you use depends on what information you have: known forces, velocity measurements, position data, or a graph.
When Gravity Is the Only Vertical Force
For any object in free fall or projectile motion, where the only force acting is gravity, the vertical acceleration is simply 9.8 m/s² directed downward. The internationally accepted standard is 9.80665 m/s², though 9.8 m/s² (or sometimes 10 m/s² for quick estimates) works for most textbook problems. This value doesn’t change based on the object’s mass, speed, or direction of travel.
In projectile motion, the key insight is that vertical and horizontal components are independent. A ball launched at an angle has zero horizontal acceleration (ignoring air resistance) but a constant vertical acceleration of 9.8 m/s² downward, whether the ball is rising, at its peak, or falling. This is true even at the very top of the arc, where vertical velocity is momentarily zero.
Using Kinematic Equations
When you know some combination of velocity, displacement, and time, you can rearrange the standard kinematic equations to solve for vertical acceleration. These equations assume acceleration is constant, which is true for most introductory physics scenarios. Replace the generic variable with the vertical direction (y) and solve for a:
Equation 1: a = (v_f − v_o) / t. Use this when you know the initial vertical velocity, final vertical velocity, and time elapsed. For example, if a ball’s vertical velocity changes from 20 m/s upward to 5 m/s upward over 1.5 seconds, the vertical acceleration is (5 − 20) / 1.5 = −10 m/s², where the negative sign indicates the acceleration is downward.
Equation 2: Rearranged from y = y_o + v_o·t + ½·a·t², you get a = 2(y − y_o − v_o·t) / t². Use this when you know displacement, initial velocity, and time but not the final velocity.
Equation 3: From v_f² − v_o² = 2a(y − y_o), you get a = (v_f² − v_o²) / 2(y − y_o). This is the time-independent equation, useful when you know both velocities and the vertical distance but not how long the motion took.
Using Newton’s Second Law
When multiple forces act on an object vertically, you find the vertical acceleration by identifying all vertical forces, calculating the net force, and dividing by mass. The core equation is a = F_net / m.
Start by drawing a free-body diagram with all vertical forces. Common ones include gravity pulling down (weight = mg), tension in a rope pulling up, the normal force from a surface, applied pushes or pulls, and air resistance opposing motion. Add up all the upward forces and subtract all the downward forces to get the net vertical force. Then divide by the object’s mass.
A classic example is elevator physics. When you stand on a scale in an elevator accelerating upward, the scale reads more than your actual weight. The scale measures the normal force N, and the relationship is N = mg + ma for upward acceleration, or N = mg − ma for downward acceleration. If a 70 kg person sees a scale reading of 756 N in an accelerating elevator, you can solve for acceleration: 756 = 70(9.8) + 70a, giving a = 1 m/s² upward. If the elevator moves at constant velocity (even if that velocity is very fast), the scale reads your normal weight because acceleration is zero.
Reading It From a Velocity-Time Graph
If you have a graph of vertical velocity plotted against time, the vertical acceleration at any point is the slope of that line. For a straight line, pick any two points, find the rise (change in velocity) and divide by the run (change in time). If the graph shows a point at (0 s, 0 m/s) and another at (5 s, 50 m/s), the acceleration is (50 − 0) / (5 − 0) = 10 m/s².
For curved velocity-time graphs, the acceleration is changing over time. The instantaneous vertical acceleration at a specific moment equals the slope of the tangent line at that point. In practice, you draw a line that just touches the curve at the moment you care about, pick two points on that tangent line, and calculate rise over run the same way.
Using Calculus for Non-Constant Acceleration
When vertical velocity isn’t changing at a steady rate, calculus gives you the exact acceleration at any instant. Vertical acceleration is the derivative of vertical velocity with respect to time: a(t) = dv/dt. If you have a position function instead, take the derivative twice, since velocity is the first derivative of position and acceleration is the second.
For example, if an object’s vertical position is described by y(t) = 4t³ − 2t + 7, its velocity is v(t) = 12t² − 2 (first derivative), and its acceleration is a(t) = 24t (second derivative). At t = 3 seconds, the vertical acceleration would be 72 m/s². This approach is essential for situations where forces change over time, such as a rocket burning fuel or a bungee cord stretching.
When Air Resistance Changes the Answer
In real-world scenarios, vertical acceleration isn’t always a constant 9.8 m/s². Air resistance creates a drag force that opposes motion, and that drag force grows with the square of the object’s speed. As a falling object speeds up, drag increases until it exactly equals the object’s weight. At that point the net vertical force is zero, so vertical acceleration drops to zero and the object falls at a constant speed called terminal velocity.
This means a skydiver’s vertical acceleration starts at 9.8 m/s² the instant they jump, then gradually decreases as they pick up speed, eventually reaching zero at terminal velocity. To find vertical acceleration at any point during this process, you return to Newton’s Second Law: a = (mg − F_drag) / m, where F_drag depends on the object’s speed, shape, and the density of the air around it. For a simple classroom problem, you’ll usually be told to ignore air resistance, but it’s worth understanding why real falling objects don’t accelerate forever.