Acceleration due to Gravity

Constant Acceleration: The acceleration due to gravity is considered constant in magnitude and direction near the Earth's surface. This constancy allows the use of the suvat equations (equations of motion for constant acceleration) to analyze vertical motion, simplifying complex trajectories into predictable kinematic relationships.

Directional Convention: Gravity always acts downwards. When applying suvat equations, it is critical to establish a consistent positive direction for displacement, velocity, and acceleration. If upwards is chosen as positive, then the acceleration 'a' in suvat equations will be $-g$. Conversely, if downwards is chosen as positive, 'a' will be $+g$. Inconsistent direction choices are a common source of error.

Vector Nature: Both displacement and velocity are vector quantities, meaning they have both magnitude and direction. The sign convention chosen for 'g' must align with the signs of initial velocity, final velocity, and displacement to ensure accurate calculations within the suvat framework.

Problem-Solving Steps: Solving freefall problems with suvat equations follows a structured approach similar to horizontal motion. First, sketch a diagram to visualize the motion and clearly define the positive direction. Next, list the known suvat variables ($s, u, v, a, t$) and the unknown variable you need to find, ensuring 'a' is correctly assigned as $+g$ or $-g$ based on your chosen positive direction.

Selecting the Correct Equation: Once three suvat variables are known, choose the appropriate suvat equation that relates these three to the unknown variable. The five suvat equations are: $v = u + at$, $s = ut + \frac{1}{2}at^2$, $s = vt - \frac{1}{2}at^2$, $s = \frac{1}{2}(u+v)t$, and $v^2 = u^2 + 2as$. Remember that these equations are only valid for constant acceleration.

Identifying Key Conditions: In vertical motion, specific conditions are often implied: at the maximum height, the instantaneous vertical velocity ($v$) is zero. If an object returns to its starting position, its displacement ($s$) is zero. These conditions provide crucial information for solving problems, often reducing the number of unknowns.

Displacement vs. Distance: In vertical motion, the distinction between displacement ($s$) and distance travelled is particularly important. Displacement is a vector quantity representing the net change in position from the starting point, so if an object returns to its origin, its displacement is zero. Distance travelled is a scalar quantity representing the total path length covered, which is always positive and accumulates regardless of direction.

Velocity at Maximum Height vs. Impact: A common point of confusion is the object's velocity at different stages. At its maximum height, an object momentarily stops before falling back down, meaning its instantaneous vertical velocity is zero ($v=0$). However, when an object hits the ground after falling, its velocity is generally not zero; it possesses a non-zero speed just before impact, which is then brought to zero by the impact force.

Deceleration vs. Negative Acceleration: When an object is moving upwards, gravity causes it to slow down, which can be described as deceleration. If the positive direction is chosen as upwards, the acceleration 'a' will be $-g$. If a question asks for the 'deceleration', the numerical value should be given as positive, as the term 'deceleration' already implies the negative direction relative to motion.

Consistent Direction: Always explicitly state your chosen positive direction (e.g., "Let upwards be positive") at the start of your solution. This helps avoid sign errors for velocity, displacement, and acceleration, and makes your working clear to the examiner.

Diagrams are Essential: Sketching a simple diagram for vertical motion problems helps visualize the initial and final positions, the direction of motion, and the constant downward acceleration due to gravity. This visual aid can prevent misinterpretations of the problem statement.

Significant Figures: When using the numerical value $g = 9.8 \text{ m/s}^2$, ensure your final answers are rounded appropriately, typically to 2 or 3 significant figures. Using more significant figures than the input values can imply a false precision and may lead to loss of marks.

Interpreting Problem Phrases: Pay close attention to keywords: "dropped" implies $u=0$. "Returns to starting position" implies $s=0$. "Maximum height" implies $v=0$. These phrases are critical for identifying known suvat variables.

Incorrect Sign for 'g': A frequent error is using a positive 'g' when the chosen positive direction is upwards, or vice-versa. Always ensure the sign of 'a' (which is 'g') is consistent with your chosen positive direction for the entire problem.

Assuming $v=0$ at Impact: Many students mistakenly assume that the final velocity ($v$) is zero when an object hits the ground. The velocity is only zero at the peak of its trajectory; it has a non-zero speed just before impact.

Confusing Displacement and Distance: Forgetting that displacement can be zero or negative while distance travelled is always positive can lead to incorrect answers, especially in problems where an object moves up and then down.

Ignoring Air Resistance: In introductory physics, problems involving gravity often assume negligible air resistance. However, in real-world scenarios, air resistance is a significant factor that would alter the acceleration and trajectory, making the constant 'g' assumption invalid.