Edexcel IAL - Mechanics 2- 1.2 Projectile Motion- Study notes - New syllabus
Edexcel IAL – Mechanics 2- 1.2 Projectile Motion -Study notes- New syllabus
Edexcel IAL – Mechanics 2- 1.2 Projectile Motion -Study notes -Edexcel A level Maths- per latest Syllabus.
Key Concepts:
- 1.2 Projectile Motion
Projectile Motion
Projectile motion occurs when a particle is projected into the air and then moves under the action of gravity only. Air resistance is neglected, so the only acceleration acting on the particle is due to gravity.
The motion takes place in a vertical plane and is analysed by resolving the motion into two perpendicular directions:
- Horizontal direction
- Vertical direction
Key Assumptions
- Acceleration due to gravity \( g \) acts vertically downwards
- Horizontal acceleration is zero
- Horizontal and vertical motions are independent
Resolution of Velocity
If a particle is projected with speed \( u \) at an angle \( \theta \) above the horizontal, the initial velocity is resolved into:
- Horizontal component: \( u\cos \theta \)
- Vertical component: \( u\sin \theta \)
Equations of Motion
Horizontal motion:
Constant velocity \( u\cos \theta \)
Displacement \( R = (u\cos \theta)t \)
Vertical motion:
Initial velocity \( u\sin \theta \)
Acceleration \( -g \)
The standard constant-acceleration equations are applied in the vertical direction.
Important Features
- At the highest point, vertical velocity is zero
- Time of flight depends only on vertical motion
- Horizontal range depends on both horizontal speed and time of flight
Projectile Motion Formula Summary
| Quantity | Formula | Notes |
|---|---|---|
| Initial velocity (horizontal) | \( u_x = u\cos \theta \) | Constant throughout motion |
| Initial velocity (vertical) | \( u_y = u\sin \theta \) | Affected by gravity |
| Horizontal acceleration | \( a_x = 0 \) | No horizontal force |
| Vertical acceleration | \( a_y = -g \) | Downward direction |
| Horizontal displacement | \( x = (u\cos \theta)t \) | Linear in time |
| Vertical displacement | \( y = (u\sin \theta)t – \dfrac{1}{2}gt^2 \) | Parabolic motion |
| Vertical velocity at time \( t \) | \( v_y = u\sin \theta – gt \) | Becomes zero at max height |
| Time to maximum height | \( t = \dfrac{u\sin \theta}{g} \) | When \( v_y = 0 \) |
| Maximum height | \( H = \dfrac{u^2\sin^2 \theta}{2g} \) | Measured from point of projection |
| Time of flight (same level) | \( T = \dfrac{2u\sin \theta}{g} \) | Launch and landing heights equal |
| Horizontal range | \( R = \dfrac{u^2\sin 2\theta}{g} \) | Maximum when \( \theta = 45^\circ \) |
Example :
A particle is projected with speed \( 20 \,\text{m s}^{-1} \) at an angle of \( 30^\circ \) above the horizontal. Find the time of flight.
▶️ Answer/Explanation
Vertical component of initial velocity:
\( u_y = 20\sin 30^\circ = 10 \)
Take upward as positive.
At landing, displacement \( s = 0 \).
Using \( s = ut + \dfrac{1}{2}at^2 \):
\( 0 = 10t – \dfrac{1}{2}gt^2 \)
\( t = \dfrac{20}{g} \)
Conclusion: The time of flight is \( \dfrac{20}{g} \) s.
Example :
A particle is projected horizontally from the top of a cliff with speed \( 12 \,\text{m s}^{-1} \). The cliff is \( 18 \) m high. Find the horizontal distance travelled before hitting the ground.
▶️ Answer/Explanation
Vertical motion determines the time of flight.
\( u_y = 0,\; s = 18,\; a = g \)
Using \( s = \dfrac{1}{2}gt^2 \):
\( 18 = \dfrac{1}{2}gt^2 \Rightarrow t^2 = \dfrac{36}{g} \)
Horizontal distance:
\( x = 12t = 12\sqrt{\dfrac{36}{g}} \)
Conclusion: The horizontal distance is \( 12\sqrt{\dfrac{36}{g}} \) m.
Example :
A particle is projected with speed \( 25 \,\text{m s}^{-1} \) at an angle of \( 45^\circ \) above the horizontal. Find the maximum height reached.
▶️ Answer/Explanation
Vertical component of initial velocity:
\( u_y = 25\sin 45^\circ = \dfrac{25}{\sqrt{2}} \)
At maximum height, \( v_y = 0 \).
Using \( v^2 = u^2 + 2as \):
\( 0 = \dfrac{625}{2} – 2gs \)
\( s = \dfrac{625}{4g} \)
Conclusion: The maximum height reached is \( \dfrac{625}{4g} \) m.