# Projectile Motion

## Projectile Motion Revision

**Projectile Motion**

Instead of objects moving in a straight line, often we observe **projectiles** and perform calculations on them. This section looks at **projectiles** in **motion** and **drag forces** that affect **projectiles**.

**Horizontal and Vertical Components of Motion**

When performing calculations on **projectiles**, it is important to look at the motion** vertically** and** horizontally** independently.

The **vertical motion** of a **projectile** determines the maximum height reached, the time in flight and therefore contributes to the overall **distance** a **projectile** flies.

The **horizontal** motion of the **projectile **purely influences the** distance** the **projectile **moves in flight.

The only **force** acting on the **projectile** is **gravity** (as most questions ask you to ignore drag forces).

Before performing calculations on **projectiles**, it is important to resolve the** projectile’s velocity** into its** horizontal** and **vertical** **components**. Trigonometry can be used to **resolve vectors**. Then we can use** SUVAT** in our calculations.

**Drag Forces**

**Drag forces** are forces that oppose the motion of an object. **Drag forces** always act in the opposite direction to motion and slow an object down. In the process, they usually convert** kinetic energy** into heat and sound. The two **drag forces** we need to discuss and learn are** friction** and **air resistance**.

**Friction** is the drag force that occurs when two surfaces rub against one another. **Friction** always increases as the speed of the object increases and is higher in heavier objects. You can reduce the effects of friction by **lubricating surfaces** or using smoother materials.

When discussing motion in a straight line on the ground,** friction **will always cause **acceleration**, **velocity** and **distance** travelled to be lower than calculated and work done to be higher than calculated.

**Air resistance** is caused when the particles of the air interact with the object moving through it. As the object collides with the air particles, it transfers some of its **kinetic energy **to the air particles as heat and sound. **Air resistance** can be reduced by **streamlining** the object, flying at a higher altitude or reducing speed.

When discussing projectile motion,** air resistance** will always cause the **vertical displacement** and **horizontal displacement** to be lower than our calculated value.

**Example: Calculations involving projectiles**

A golf ball is hit so that it leaves the ground at an angle of 30° to the horizontal. It leaves the floor with a velocity of 35 \: \text{ms}^{-1}. Calculate how far the ball will travel before hitting the ground. Ignore the effects of drag forces.

**[5 marks]**

**Step 1:** Resolve the velocity into its horizontal and vertical components.

\text{Horizontal} = \textcolor{bd0000}{35} \cos \textcolor{7cb447}{30 \degree} = \boldsymbol{30.31} \: \textbf{ms}^{-1}

\text{Vertical} = \textcolor{bd0000}{35} \sin \textcolor{7cb447}{30 \degree} = \boldsymbol{17.5} \: \textbf{ms}^{-1}

**Step 2:** Determine the time in flight using SUVAT.

Time of flight is determined by vertical motion only. Therefore we only need to consider the vertical motion of the golf ball.

\begin{aligned}S &= \: ? \\ U &= 17.5 \: \text{ms}^{-1} \\ V &= 0 \\ A &= -9.81 \: \text{ms}^{-2} \\ T &= \: ? \end{aligned}

We know V = 0 because at the top of the flight, the vertical velocity is momentarily 0 \: \text{ms}^{-1}.

The acceleration is negative as it is opposing the motion.

We want to find t, the time. We can use v = u + at in this instance.

0 = 17.5 + \left(-9.81 \times t\right)

Rearranging this gives:

9.81 \times t = 17.5

t = \dfrac{17.5}{9.81} = \boldsymbol{1.784} \: \textbf{s}

This is only the time to reach max height so we need to double this for the falling down part of the flight. Therefore:

t = 1.784 \times 2 = \boldsymbol{3.568} \: \textbf{s}

**Step 3:** Using our time and horizontal velocity, we can determine distance using v = \dfrac{s}{t}.

s = v \times t

s = 30.31 \times 3.568 = 108.14 \: \text{m}

s = \boldsymbol{110} \: \textbf{m}

## Projectile Motion Example Questions

**Question 1:** What forces are acting on an object in projectile motion when drag forces are negligible?

**[1 mark]**

The only force acting on an object in projectile motion is** gravity** (or the object’s own **weight**).

Drag forces are considered negligible for projectile motion questions.

**Question 2:** A golf ball is hit so that it leaves the ground at an angle of 22 \degree to the horizontal. It leaves the floor with a velocity of 75 \: \text{ms}^{-1}. Calculate the maximum height of the ball. Ignore the effects of drag forces.

**[3 marks]**

S= \: ? (This is what we are looking)

U= 28.1 \: \text{ms}^{-1}

V= \: 0 \: \text{ms}^{-1} (at the top of the flight, the vertical velocity is momentarily 0 \: \text{ms}^{-1})

A= -9.81 \: \text{ms}^{-2} (minus as the acceleration is opposing motion)

T= \: ? (not needed)

v^2 = u^2 + 2as

u = 0 \: \text{ms}^{-1} so v^2 = 2as .

\begin{aligned} s &= \dfrac{v^2}{2a} \\ \\ s &= \dfrac{28.1^2}{2 \times -9.81} \\ \\ s &= \dfrac{789.61}{-19.62} = \boldsymbol{40.2} \: \textbf{m} \end{aligned}**Question 3:** What effect do drag forces have on projectiles compared to our calculated theoretical answers?

**[2 marks]**

Drag forces will always** reduce our values** from our calculations. D**istance, and height** will all be lower when drag forces are accounted for.

## Projectile Motion Worksheet and Example Questions

### SUVAT Questions

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