Mechanics: Projectiles and Further Kinematics
Projectiles and further kinematics is the A Level Mechanics topic covering particles moving freely under gravity in two dimensions, plus motion described by vectors or calculus where velocity and acceleration vary with time. It extends the suvat equations to horizontal and vertical components and links position, velocity and acceleration through differentiation and integration.
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Method
- For horizontal projectiles, split the motion into independent horizontal (constant velocity) and vertical (constant acceleration g) components and treat each with the suvat equations separately.
- For projectiles launched at an angle, resolve the initial velocity into horizontal (u cos theta) and vertical (u sin theta) components before applying suvat to each direction.
- Use the vertical motion to find the time of flight, for example from s = ut + 0.5at^2, then substitute that time into the horizontal equation to find range or position.
- For vector-based kinematics, differentiate the position vector with respect to time to get velocity, and differentiate velocity to get acceleration; integrate the other way, adding a constant vector found from the given initial conditions.
- To find when a particle moves parallel to a given vector, set one component of velocity to zero (parallel to i or j) or set the ratio of the components equal to the ratio of the components of that vector.
- Combine horizontal and vertical, or i- and j-, components with Pythagoras to find speed, and use tan to find the direction of motion.
Worked example
A ball is thrown horizontally with speed 12 m/s from the top of a vertical cliff 19.6 m above the sea, modelled as a horizontal plane. The ball is modelled as a particle moving freely under gravity. Take g = 9.8 m/s^2. Find (a) the time taken for the ball to reach the sea, (b) the horizontal distance travelled, and (c) the speed of the ball as it hits the sea, giving your answer to 3 significant figures.
- Vertical motion: use s = ut + 0.5at^2 with u = 0 (no initial vertical speed), a = 9.8, s = 19.6: 19.6 = 4.9t^2.
- Solve: t^2 = 4, so t = 2 seconds.
- Horizontal motion: horizontal distance = horizontal speed x time = 12 x 2 = 24 m.
- Vertical speed at impact: v = u + at = 0 + 9.8 x 2 = 19.6 m/s.
- Combine components: speed = sqrt(12^2 + 19.6^2) = sqrt(528.16) = 23.0 m/s (3 s.f.).
- Final answer: time = 2 s, horizontal distance = 24 m, impact speed = 23.0 m/s.
Practice questions
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Q1A ball is thrown horizontally with speed 8 m/s from a point 4.9 m above horizontal ground. Take g = 9.8 m/s^2. Find the time taken for the ball to reach the ground.Show answer
Answer: t = 1 s (4.9 = 4.9t^2)
Q2A particle is thrown horizontally with speed 6 m/s from a point 19.6 m above horizontal ground, modelled as a plane. Take g = 9.8 m/s^2. Find the horizontal distance travelled before it lands.Show answer
Answer: 12 m (t = 2 s from 19.6 = 4.9t^2, then 6 x 2)
Q3A particle is projected from level ground with speed 21 m/s at 40 degrees above the horizontal. Take g = 9.8 m/s^2. Find the greatest height reached, giving your answer to 3 significant figures.Show answer
Answer: 9.30 m (u_y = 21 sin40 = 13.5, H = u_y^2/(2 x 9.8))
Q4A particle P has position vector (3i - j) m at time t = 0 and moves with constant velocity (2i + 4j) m/s. Find the position vector of P at time t = 2.5 seconds.Show answer
Answer: (8i + 9j) m (r = r0 + vt)
Q5A particle is projected from horizontal ground with speed 24 m/s at an angle of 35 degrees above the horizontal. Take g = 9.8 m/s^2. Find the horizontal range, giving your answer to 3 significant figures.Show answer
Answer: 55.2 m (R = u^2 sin(2 x 35)/g = 576 x sin70 / 9.8)
Q6A particle P moves along the x-axis such that its velocity at time t seconds (0 <= t <= 5) is v = 6t - t^2 m/s, where v >= 0 throughout this interval. Given that P starts at the origin, find the distance travelled by P between t = 0 and t = 5.Show answer
Answer: 100/3 m = 33.3 m (integrate 6t - t^2 from 0 to 5: [3t^2 - t^3/3])
Exam-style questions
Written in the style of a A Level Maths exam paper, with a full mark scheme.
A stone is thrown horizontally with speed 18 m/s from the top of a vertical cliff. The point of projection is 78.4 m above the sea, modelled as a horizontal plane. Take g = 9.8 m/s^2. Find the speed of the stone as it hits the sea, giving your answer to 3 significant figures.
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A particle is projected from a point on horizontal ground with speed 35 m/s at an angle of 28 degrees above the horizontal. The particle moves freely under gravity until it returns to the ground. Take g = 9.8 m/s^2. Find (a) the time of flight and (b) the horizontal range, giving each answer to 3 significant figures.
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A particle P moves in a plane such that at time t seconds (t >= 0), its velocity is v = (4t - 2)i + (3t^2 - 1)j m/s. When t = 0, P is at the point with position vector (i + 4j) m relative to a fixed origin O. Find (a) the acceleration of P when t = 2 seconds, and (b) the position vector of P when t = 2 seconds.
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Free printable worksheet
Want more practice on paper? Download the mechanics: projectiles and further kinematics worksheet pack - 9 pages of exam-style questions with a full mark scheme. One email opens every download in this browser for 14 days - no account, no card. Print it for personal and classroom use.
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