Mechanics: Projectiles and Applications Depth
In A-level Mechanics, projectiles at depth go beyond a single horizontal or angled launch on level ground: questions launch from a height at an angle (needing a quadratic in t), ask which angle hits a given target point (using the trajectory equation), or land on a slope. The marks reward resolving the initial velocity into horizontal and vertical components before anything else, keeping the two directions independent except for a shared time, and recognising that the standard time-of-flight and range formulae only apply when the landing height equals the launch height.
Before you start
Make sure you're comfortable with these topics first:
Method
- Resolve the initial velocity into horizontal (U cos(theta), constant throughout, since there is no horizontal force) and vertical (U sin(theta), changing under gravity) components before doing anything else.
- Treat horizontal and vertical motion as two independent suvat problems, linked only by the shared value of time t.
- If launch and landing heights are equal, use the vertical suvat with total vertical displacement 0 (or the standard result T = 2U sin(theta)/g for the time of flight); if the heights differ, use the actual (positive or negative) vertical displacement in s = ut + 0.5at^2 and solve the resulting quadratic in t, discarding any negative root.
- To find the range, substitute the time of flight into the horizontal equation (range = horizontal speed x time); to find the greatest height, use v^2 = u^2 + 2as in the vertical direction only, with v = 0.
- For 'find the angle to hit a given point' problems, substitute the coordinates of the point into the trajectory equation y = x tan(theta) - [g x^2 (1 + tan^2(theta))]/(2U^2), which comes from eliminating t and using sec^2(theta) = 1 + tan^2(theta), then solve the resulting equation (often a quadratic in tan(theta)).
- For a projectile landing on a slope, write the slope's equation (y = x tan(alpha) for a slope rising at angle alpha, or y = -x tan(alpha) for one falling away) and find where the trajectory equation meets it; the distance along the slope is then the horizontal distance divided by cos(alpha).
- State the standard modelling assumptions (no air resistance, the object is a particle) where a question asks you to comment on the model, and keep working symbolic (in U and theta) for as long as possible in a 'show that' question.
Worked example
A particle is projected from a point O with speed 25 m/s at an angle of 30 degrees above the horizontal. The point O is 15 m above horizontal ground. The particle moves freely under gravity until it hits the ground. Take g = 9.8 m/s^2. Find (a) the time taken for the particle to hit the ground, giving your answer to 3 significant figures, (b) the horizontal distance from O to the point where the particle lands.
- Resolve the initial velocity: horizontal u_x = 25cos30 = 21.651 m/s (3 d.p.); vertical u_y = 25sin30 = 12.5 m/s.
- Taking upward as positive, the particle lands 15 m below O, so its vertical displacement is -15 m. Using s = ut + 0.5at^2 with a = -9.8: -15 = 12.5t - 4.9t^2.
- Rearrange into the standard quadratic form: 4.9t^2 - 12.5t - 15 = 0.
- Solve using the quadratic formula: t = [12.5 +- sqrt(12.5^2 + 4(4.9)(15))]/(2 x 4.9) = [12.5 +- sqrt(450.25)]/9.8.
- sqrt(450.25) = 21.219 (3 d.p.), so t = (12.5 + 21.219)/9.8 = 3.44 s (3 s.f.), rejecting the negative root.
- Horizontal distance = u_x x t = 21.651 x 3.4407 = 74.5 m (3 s.f.).
Practice questions
Try each question, then tap to reveal the answer.
Q1State which component of a projectile's velocity is constant during the flight, and which changes at a constant rate, giving a brief reason for each.Show answer
Answer: The horizontal component of velocity is constant, since there is no horizontal force acting once the particle is in flight (air resistance is ignored). The vertical component changes at a constant rate g, since gravity is the only force acting and it is vertical, giving constant (uniform) vertical acceleration.
Q2A particle is projected from horizontal ground at 50 degrees above the horizontal, and has time of flight 4 seconds before it returns to the ground. Take g = 9.8 m/s^2. Find the initial speed of projection, giving your answer to 3 significant figures.Show answer
Answer: 25.6 m/s (3 s.f.), from rearranging T = 2U sin(theta)/g to U = Tg/(2sin(theta)) = 4(9.8)/(2sin50).
Q3A ball is thrown horizontally from a point above horizontal ground and takes 3 seconds to reach the ground. Take g = 9.8 m/s^2. Find the height of the point of projection above the ground.Show answer
Answer: 44.1 m (exact), from H = 4.9t^2 = 4.9(3)^2.
Q4Two particles are projected from horizontal ground with the same speed of 20 m/s, one at 30 degrees above the horizontal and the other at 60 degrees above the horizontal. Take g = 9.8 m/s^2. Find the difference between the greatest heights they reach.Show answer
Answer: 10.2 m (3 s.f.). Greatest heights are (20sin30)^2/19.6 = 5.10 m and (20sin60)^2/19.6 = 15.3 m; note these two heights are different even though 30 and 60 degrees are complementary angles that would give the same range.
Q5A particle is projected from horizontal ground at 35 degrees above the horizontal and lands 60 m from its point of projection. Take g = 9.8 m/s^2. Find the initial speed of projection, giving your answer to 3 significant figures.Show answer
Answer: 25.0 m/s (3 s.f.), from rearranging R = U^2 sin(2theta)/g to U = sqrt(Rg/sin(2theta)) = sqrt(60(9.8)/sin70).
Q6Explain why the formulae T = 2U sin(theta)/g and R = U^2 sin(2theta)/g cannot be used directly for a projectile that lands at a different height from the one it was launched from.Show answer
Answer: Both formulae are derived by setting the total vertical displacement to zero, i.e. assuming the particle returns to its starting height. If the landing point is higher or lower than the point of projection, the vertical displacement at landing is not zero, so the time of flight must instead be found by solving s = U sin(theta) t - 0.5gt^2 with s equal to the actual (nonzero) height difference, which generally gives a quadratic in t that does not simplify to the standard formula.
Q7A particle is projected from horizontal ground with speed U at an angle theta above the horizontal (0 < theta < 90, theta not equal to 45) and has horizontal range R. State, in terms of theta, the other angle of projection using the same speed U that gives the same range R, and explain why.Show answer
Answer: 90 - theta (the complementary angle). Range = U^2 sin(2theta)/g depends on theta only through sin(2theta), and sin(2(90 - theta)) = sin(180 - 2theta) = sin(2theta), so theta and 90 - theta give the same value of sin(2theta) and hence the same range.
Exam-style questions
Written in the style of a A Level Maths exam paper, with a full mark scheme.
A particle is projected from a point O on horizontal ground with speed 20 m/s, at an angle theta above the horizontal (0 < theta < 90). The particle passes through the point which is 8 m horizontally and 3 m vertically from O. Take g = 9.8 m/s^2. Using the trajectory equation y = x tan(theta) - [g x^2 (1 + tan^2(theta))]/(2U^2), show that 98 tan^2(theta) - 1000 tan(theta) + 473 = 0, and hence find the two possible values of theta, giving your answers to 1 decimal place.
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A particle P is projected from a point on horizontal ground with velocity (9i + 15.68j) m/s, where i and j are horizontal and vertically upward unit vectors respectively. Take g = 9.8 m/s^2. Find (a) the greatest height reached by P above the ground, (3) (b) the time of flight (the time until P returns to the ground), (2) (c) the velocity of P, in the form pi + qj, at the instant it lands. (2)
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A particle is projected from the top of a slope, O, with speed 14 m/s at 40 degrees above the horizontal. The slope falls away from O at a constant angle of 15 degrees below the horizontal. The particle moves freely under gravity and lands on the slope at the point Q. Take g = 9.8 m/s^2. Using x for the horizontal distance from O to the landing point, (a) find the value of x by setting the trajectory equation equal to the equation of the slope. (5) (b) Hence find the distance OQ, giving your answer to 3 significant figures. (2)
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