Projectiles (Leaving Cert Physics): Revision Notes

Projectiles

What is projectile motion?

Projectile motion occurs when an object is thrown or launched into the air and follows a curved path under the influence of gravity alone. Any object that moves through the air in this way is called a projectile. Examples include a football being kicked, a basketball being shot, or a stone being thrown.

Key principles of projectile motion

Horizontal motion characteristics

For the horizontal direction:

• No forces act horizontally (ignoring air resistance)
• The horizontal velocity remains constant throughout the entire flight
• Horizontal acceleration = 0
• Distance = velocity × time

Vertical motion characteristics

For the vertical direction:

• Gravity acts downward with acceleration $g = 9.8 \text{ m s}^{-2}$
• The vertical velocity changes continuously due to gravity
• We use the standard kinematic equations with $a = -9.8 \text{ m s}^{-2}$ (negative because gravity acts downward)

Trajectory characteristics

All projectiles follow a parabolic path when launched at an angle. The shape of this curve depends on the initial velocity and launch angle.

Problem-solving approach

Step 1: Resolve the initial velocity

When a projectile is launched at an angle $θ$ with initial speed $u$, we must find the horizontal and vertical components:

• Horizontal component: $u_x = u \cos θ$
• Vertical component: $u_y = u \sin θ$

Step 2: Apply equations separately

For horizontal motion (acceleration = 0):

• $v_x = u_x$ (velocity stays constant)
• $s_x = u_x t$ (distance = velocity × time)

For vertical motion (acceleration = $-9.8 \text{ m s}^{-2}$):

• $v_y = u_y - gt$
• $v_y^2 = u_y^2 - 2gs_y$
• $s_y = u_y t - \frac{1}{2}gt^2$

Step 3: Combine results when needed

To find the resultant velocity at any time:

• Magnitude: $v = \sqrt{v_x^2 + v_y^2}$
• Direction: $\tan θ = \frac{v_y}{v_x}$

Common calculations

Time of flight

For a projectile that lands at the same height it was launched from:

• At landing, the vertical displacement is zero
• Use: $0 = u_y t - \frac{1}{2}gt^2$
• This gives: $t = \frac{2u_y}{g}$ (ignoring $t = 0$ at launch)

Maximum height

The highest point is reached when the vertical velocity becomes zero:

• Use: $v_y^2 = u_y^2 - 2gh_{max}$
• At maximum height: $v_y = 0$
• Therefore: $h_{max} = \frac{u_y^2}{2g}$

Horizontal range

The total horizontal distance travelled:

• Range = horizontal velocity × time of flight
• $R = u_x × \frac{2u_y}{g}$

Worked example method