Motion Under Gravity Revision Notes for Leaving Cert Applied Maths

Motion Under Gravity

Understanding gravity in physics

You may remember studying gravity during Junior Science, where you learned about calculating mass and weight. In those calculations, we often used a simplified value of 10 m/s² for gravity. However, the actual acceleration due to gravity is approximately 9.8 m/s², and this is the value we use in Applied Maths.

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The transition from the simplified gravity value of 10 m/s² in Junior Science to the more precise 9.8 m/s² in Applied Maths reflects the need for greater accuracy in advanced physics calculations.

When objects fall or are thrown, they experience this constant acceleration of 9.8 m/s² towards the Earth. This makes motion under gravity another excellent example of uniform acceleration, which means we can apply all the equations of motion we've studied previously.

Sign conventions for gravitational acceleration

The direction we choose as positive is crucial when solving gravity problems. Understanding and correctly applying sign conventions is one of the most important skills in gravitational motion problems.

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Critical Sign Convention Rules:

Falling objects: Acceleration = +g (+9.8 m/s²)
Objects thrown upwards: Acceleration = -g (-9.8 m/s²)

This sign convention reflects that gravity always acts downwards. When an object is thrown upwards, gravity opposes the motion, creating a negative acceleration that gradually reduces the upward velocity until it reaches zero at the maximum height.

Key equations for motion under gravity

Since gravitational motion involves uniform acceleration, we use the same kinematic equations. The key is correctly identifying the acceleration value: $a = +9.8 \text{ m/s}^2$ for falling objects and $a = -9.8 \text{ m/s}^2$ for upward motion.

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The three fundamental kinematic equations for motion under gravity:

$s = ut + \frac{1}{2}at^2$ (displacement equation)
$v^2 = u^2 + 2as$ (velocity-displacement equation)
$v = u + at$ (velocity-time equation)

Where $a = \pm 9.8 \text{ m/s}^2$ depending on the direction of motion.

Worked example: Stone thrown vertically upward

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Worked Example: Stone Thrown Vertically Upward

Problem: A stone is thrown vertically upwards from ground level with an initial speed of 35 m/s. i) After how long will the stone hit the ground? ii) Find the greatest height reached.

Part i: Finding time to hit the ground

Given values:

Initial velocity: $u = 35 \text{ m/s}$ (upward)
Acceleration: $a = -9.8 \text{ m/s}^2$ (gravity opposes upward motion)
Final displacement: $s = 0$ (returns to ground level)
Time: $t = ?$ (what we're finding)

Solution approach: We use the displacement equation: $s = ut + \frac{1}{2}at^2$

Substituting our values: $0 = (35)t + \frac{1}{2}(-9.8)t^2$ $0 = 35t - 4.9t^2$ $4.9t^2 - 35t = 0$ $t(4.9t - 35) = 0$

This gives us: $t = 0$ or $4.9t - 35 = 0$

The solution $t = 0$ represents the starting point. The meaningful answer is: $t = \frac{35}{4.9} = \frac{50}{7} = \text{**7.1 seconds**}$

Part ii: Finding maximum height

At the maximum height, the stone's velocity becomes zero before it starts falling back down.

Given values:

Initial velocity: $u = 35 \text{ m/s}$
Acceleration: $a = -9.8 \text{ m/s}^2$
Final velocity: $v = 0 \text{ m/s}$ (at maximum height)
Displacement: $s = ?$ (maximum height)

Solution approach: We use the velocity-displacement equation: $v^2 = u^2 + 2as$

Substituting our values: $(0)^2 = (35)^2 + 2(-9.8)s$ $0 = 1225 - 19.6s$ $19.6s = 1225$ $s = \frac{1225}{19.6} = \text{**62.5 metres**}$

Important concepts to remember

Understanding the physics behind gravitational motion helps us solve problems more effectively and avoid common mistakes.

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Maximum height characteristics:

At maximum height, the velocity is always zero
The time to reach maximum height equals the time to fall back down from that height
This symmetry is a key feature of projectile motion under gravity

Problem-solving strategy:

Identify what you're looking for (time, height, velocity)
Choose the appropriate kinematic equation
Apply the correct sign for acceleration (+g for falling, -g for upward motion)
Substitute values carefully and solve

Essential reminders for motion under gravity

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Key Points to Remember:

Gravity causes a constant acceleration of 9.8 m/s² towards Earth
Use negative acceleration (-9.8 m/s²) when objects move upward against gravity
At maximum height, velocity equals zero - this is often a key condition in problems
The time to reach maximum height equals the time to return from maximum height
Motion under gravity follows the same kinematic equations as any uniform acceleration

Motion Under Gravity (Leaving Cert Applied Maths): Revision Notes