Projectile
Motion & Kinematics - OCR A-Level Physics
Key Definition
Projectile
An object launched into the air that moves freely under gravity alone (ignoring air resistance). Its trajectory is a parabola.
An object launched into the air that moves freely under gravity alone (ignoring air resistance). Its trajectory is a parabola.
- The key principle: horizontal and vertical components of motion are independent of each other
- Horizontally: no accelerationThe rate of change of velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹.. A vector quantity. Measured in m s⁻². (ignoring air resistanceThe opposition to currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A). flow. The ratio of potential difference to currentThe rate of flow of chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C).. Measured in amperes (A).. Measured in ohms (Ω).), so v_x is constant and $s_{x} = v_{x} t$
- Vertically: accelerationThe rate of change of velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹.. A vector quantity. Measured in m s⁻². = g = 9.81 $m s^{-2}$ downwards. Use SUVAT equations with a = g
- The time of flight is determined by the vertical motion; this same time is used for horizontal calculations
- Resolve the launch velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹. into components: v_x = $v\cos\theta$ and v_y = $v\sin\theta$
- At the highest point, the vertical component of velocity is zero but the horizontal component is unchanged
- The trajectory is symmetrical: the time to reach maximum height equals the time to fall back to the same level
$$v_{x} = v\cos\theta$$
$$v_{y} = v\sin\theta$$
Common Mistake
MEDIUM
Students often: A common slip: using the full launch speed in both horizontal and vertical SUVAT equations instead of resolving into components.
Instead: Resolve the initial velocity into horizontal (v cos theta) and vertical (v sin theta) components first. For a horizontal launch, the initial vertical velocity is zero.
Instead: Resolve the initial velocity into horizontal (v cos theta) and vertical (v sin theta) components first. For a horizontal launch, the initial vertical velocity is zero.
Worked Example [4 marks]
A ball is kicked horizontally from the top of a cliff at 15 $m s^{-1}$. The cliff is 45 m high. Calculate the time to reach the ground and the horizontal distance travelled. Use g = 9.81 $m s^{-2}$.
Show Solution
1
Vertical
$s = 45 \text{ m}, u = 0 (horizontal launch), a = 9.81 \text{ m s}^{-2}$
[1]2
$s = ut + \frac{1}{2}at^{2} \Rightarrow 45 = \frac{1}{2}(9.81)t^{2} \Rightarrow t = \sqrt{\frac{90}{9.81}} = 3.03 \text{ s}$
[1]3
Horizontal
$no accelerationThe rate of change of velocity. A vector quantity. Measured in m s⁻²., so range = v_x \times t = 15 \times 3.03 = 45.4 \text{ m}$
[1]4
Time to reach ground = 3.0 s; horizontal $distance = 45 m (2 s.f.)$
[1]Answer
Time = 3.0 s; horizontal distance = 45 m