Modelling a projectile as a particle moving in a vertical plane with constant acceleration g = 9.8 m s⁻² downwards

Horizontal and vertical components of the initial velocity and of the motion

Time of flight, range and greatest height, and deriving the general results

Speed and direction of motion at a given time

Projection from a height, and projection to pass through or hit a given point

Finding the angle of projection: a quadratic in tan θ using sec²θ= 1 + tan²θ

The equation of the trajectory, and position and velocity vectors in terms of i and j

1. A projectile's horizontal and vertical motions are independent: horizontally the velocity never changes, vertically there is a constant acceleration g downwards, and time is the only thing the two directions share.

2. Split the initial velocity: horizontal ucos θ, vertical usin θ.

3. Horizontally x = (ucos θ)t; vertically use the constant-acceleration formulae with a = -9.8 (upwards positive).

4. On level ground: time of flight 2usin θ/g, range u²sin 2θ/g, greatest height u²sin²θ/(2g). None is in the formulae booklet: derive them.

5. Velocity at time t: combine the components with Pythagoras and tan, and say whether the direction is above or below the horizontal.

6. Trajectory: y = xtan θ- gx²/2u²cos²θ. Writing 1/cos²θ= 1 + tan²θ gives a quadratic in tan θ.

The modelling assumptions are: the body is a particle (size and spin ignored); there is no air resistance, so the only force after launch is the weight mg; and g is constant at 9.8 m s⁻². So the acceleration is g vertically downwards, the same for every mass, and the path lies in the vertical plane containing the initial velocity.

A particle projected with speed u at angle θ above the horizontal (the angle of projection) has initial velocity components ucos θ horizontally and usin θ vertically.

The vertical column is just the M1 constant-acceleration formulae with a = -g; v² = u² + 2as and s = ½(u + v)t are available too. Time links the two directions: find t from one direction, then use it in the other.

At the greatest height the vertical velocity is zero: use v² = u² + 2as vertically with v = 0. For the time of flight back to launch level, the vertical displacement is zero: use s = ut + ½at² with s = 0 and take the non-zero root. The range is the horizontal distance travelled in that time. On level ground the path is symmetrical, so the time to the top is half the time of flight.

WORKED EXAMPLE · QUESTION

A football is kicked from a point O on horizontal ground with speed 18 m s⁻¹ at 35° above the horizontal. The ball is modelled as a particle moving freely under gravity. Find (a) the greatest height of the ball above the ground, (b) the time for which the ball is in the air, (c) the horizontal distance from O to the point where the ball first lands.

SOLUTION

Components: horizontal 18cos 35° = 14.745 m s⁻¹, vertical 18sin 35° = 10.324 m s⁻¹.

(a) At the top v = 0:

0 = 10.324² - 2 ×9.8 ×h, so h = 5.438. 5.44 m

(b) Vertically s = 0 when it lands: 0 = 10.324t - 4.9t² = t(10.324 - 4.9t), so t = 10.324 ÷4.9 = 2.107. 2.11 s

(c) Horizontally: x = 14.745 ×2.107 = 31.07. 31.1 m

Check with symmetry: time to the top = 10.324 ÷9.8 = 1.054 s, exactly half of 2.107 s.

The general results (level ground)

From 0 = (usin θ)t - ½gt², the non-zero root is T = 2usin θ/g. Then R = ucos θ×2usin θ/g = u²sin 2θ/g, using the double-angle identity from P3. From 0 = (usin θ)² - 2gH, H = u²sin²θ/2g. None of the three is in the formulae booklet, and they are valid only when the particle lands at the level it was launched from; a "show that" needs the derivation.

For a fixed speed, sin 2θ is greatest when 2θ= 90°, so the maximum range is u²/g, at θ= 45°. Since sin 2θ= sin(180° - 2θ), the angles θ and 90° - θ give the same range, with different heights and times.

WORKED EXAMPLE · QUESTION

A sprinkler at ground level projects water drops with speed 9.0 m s⁻¹. The angle of projection can be adjusted. The drops are modelled as particles and the lawn is horizontal. (a) Show that the range R m at angle θ is given by R = 81sin 2θ/9.8. (b) Find the greatest possible range. (c) Find the two angles of projection that give a range of 7.0 m.

SOLUTION

(a) Vertically 0 = (9sin θ)t - 4.9t², so t = 9sin θ/4.9. Then R = 9cos θ×9sin θ/4.9 = (81 ×2sin θcos θ)/9.8 = 81sin 2θ/9.8, as required.

(b) The greatest value of sin 2θ is 1, at θ= 45°: R_(max) = 81 ÷9.8. 8.27 m

(c) sin 2θ= 7.0 ×9.8 ÷81 = 0.8469, so 2θ= 57.88° or 122.12°. θ = 28.9° or 61.1°

The two answers add to 90°, as they must.

The horizontal component is vₓ = ucos θ (unchanged) and the vertical component is v_y = usin θ- gt: positive means moving upwards, negative means moving downwards. The speed is √(vₓ² + v_y²), and the direction makes angle α with the horizontal where tan α= |v_y| ÷vₓ, stated as above or below. The velocity is always along the tangent to the path. To find a speed at a given height without the time, use v² = u² + 2as vertically, then combine with the horizontal component.

WORKED EXAMPLE · QUESTION

A particle P is projected from a point on horizontal ground with speed 25 m s⁻¹ at an angle of 50° above the horizontal. (a) Find the speed of P and its direction of motion 3.0 s after projection. (b) Find the two times at which P is moving in a direction at 20° to the horizontal.

SOLUTION

Components: 25cos 50° = 16.070 m s⁻¹ and 25sin 50° = 19.151 m s⁻¹.

(a) v_y = 19.151 - 9.8 ×3.0 = -10.249 m s⁻¹ (moving downwards).

Speed = √(16.070² + 10.249²) = 19.06 and tan α= 10.249 ÷16.070 gives α= 32.5°. 19.1 m s⁻¹ at 32.5° below the horizontal

(b) At 20°, |v_y| = 16.070tan 20° = 5.849 m s⁻¹. Moving up: 19.151 - 9.8t = 5.849, t = 1.36. Moving down: 19.151 - 9.8t = -5.849, t = 2.55. t = 1.36 s and 2.55 s

The two times are symmetrical about the time of the top, 19.151 ÷9.8 = 1.95 s.

Take the point of projection as the origin with upwards positive; landing h metres below the launch point means vertical displacement -h. The time comes from a quadratic, -h = (usin θ)t - 4.9t²; reject the negative root, which describes a time before the launch. The range formula does not apply, because launch and landing are at different levels. Projected horizontally, the initial vertical velocity is zero and -h = -4.9t² gives the time directly; projected below the horizontal, the initial vertical component is -usin θ.

WORKED EXAMPLE · QUESTION

A stone is thrown from a point O at the top of a vertical cliff. O is 40 m above the sea. The stone is thrown with speed 15 m s⁻¹ at 25° above the horizontal and moves freely under gravity until it hits the sea. Find (a) the time taken for the stone to reach the sea, (b) the horizontal distance from the foot of the cliff to the point where the stone hits the sea, (c) the speed of the stone as it hits the sea and the angle its path then makes with the horizontal.

SOLUTION

Components: 15cos 25° = 13.595 m s⁻¹ and 15sin 25° = 6.339 m s⁻¹.

(a) Vertically, upwards positive, s = -40: 4.9t² - 6.339t - 40 = 0, so t = (6.339 ±√(6.339² + 4 ×4.9 ×40))/9.8 = 3.576 (the other root, -2.28, is rejected). 3.58 s

(b) x = 13.595 ×3.576 = 48.62. 48.6 m

(c) v_y = 6.339 - 9.8 ×3.576 = -28.71 m s⁻¹. Speed = √(13.595² + 28.71²) = 31.76, and tan α= 28.71 ÷13.595 gives α= 64.7°. 31.8 m s⁻¹ at 64.7° below the horizontal

Check without the time: v² = 15² + 2 ×9.8 ×40 = 1009, so v = 31.76 m s⁻¹; the speed depends only on the drop, not on the angle.

A target a m horizontally and b m vertically from the point of projection gives two conditions: x = a and y = b at the same time. With a known velocity, find t from the horizontal motion and substitute in the vertical equation. With an unknown speed or angle, eliminate t using t = a ÷(ucos θ); this is where the trajectory equation comes from. The sign of v_y at that time tells you whether the particle is rising or falling, and to check whether it clears an obstacle, compare its height at the obstacle's horizontal distance with the obstacle's height.

WORKED EXAMPLE · QUESTION

A ball is thrown from a point A, 1.5 m above horizontal ground, towards a vertical wall. The wall is 25 m horizontally from A. The ball is thrown with speed 20 m s⁻¹ at 30° above the horizontal, in a vertical plane perpendicular to the wall. (a) Find the time taken for the ball to reach the wall. (b) Find the height above the ground at which the ball hits the wall. (c) Determine whether the ball is rising or falling when it hits the wall, and find its speed at that instant.

SOLUTION

(a) Horizontally, 20cos 30° = 17.321 m s⁻¹, so t = 25 ÷17.321 = 1.4434. 1.44 s

(b) Vertically, initial component 20sin 30° = 10 m s⁻¹: y = 10 ×1.4434 - 4.9 ×1.4434² = 4.225 m above A, so 4.225 + 1.5 above the ground. 5.73 m

(c) v_y = 10 - 9.8 ×1.4434 = -4.15 m s⁻¹, negative, so the ball is falling. Speed = √(17.321² + 4.15²) = 17.81. falling, 17.8 m s⁻¹

Keep t = 1.4434 s unrounded in later parts; rounding it to 1.44 first would shift the height in the third significant figure.

Eliminate t between the two displacements. From x = (ucos θ)t, t = x ÷(ucos θ); substitute into y = (usin θ)t - ½gt²:

The origin is the point of projection, with y vertically upwards; the second form uses 1/cos²θ= sec²θ= 1 + tan²θ. It is not in the formulae booklet, and questions often ask you to show it. For fixed u and θ this is y = px - qx²: the path is a parabola.

WORKED EXAMPLE · QUESTION

A tennis ball is struck from a point O with speed 14 m s⁻¹ at an angle θ above the horizontal, where tan θ= ¾. Take O as the origin, with x and y measured horizontally and vertically upwards in metres. (a) Show that the equation of the path is y = ¾x - (5/128)x². (b) Use the equation to find the range on the horizontal plane through O and the greatest height. (c) Determine whether the ball passes over a net 1.8 m high at a horizontal distance of 16 m from O.

SOLUTION

(a) tan θ= ¾ gives sin θ= 0.6, cos θ= 0.8. Then y = ¾x - 9.8x²/(2 ×14² ×0.8²) = ¾x - (9.8/250.88)x² = ¾x - (5/128)x², as required.

(b) Range: y = 0, x ≠0: x = ¾ ×(128/5) = 19.2. The greatest height is at the midpoint, x = 9.6: y = 7.2 - 3.6 = 3.6. range 19.2 m, greatest height 3.6 m

(c) At x = 16: y = 12 - (5/128) ×256 = 12 - 10 = 2.0 m > 1.8 m. It passes over the net, by 0.2 m

Check (b) with the general results: 14² ×2(0.6)(0.8) ÷9.8 = 19.2 and (14 ×0.6)² ÷19.6 = 3.6.

If the speed u is known and the path must pass through a point (a, b), substitute into the second form of the trajectory equation. It becomes a quadratic in tan θ:

Two positive roots give two possible angles: a low, fast-arriving path and a high, slow "lob". One repeated root means the point is just within reach at that speed; no real roots means it cannot be reached at speed u. Derive this from the trajectory equation each time; learn the identity sec²θ= 1 + tan²θ.

WORKED EXAMPLE · QUESTION

A particle is projected from a point O with speed 28 m s⁻¹ at an angle θ above the horizontal. It passes through the point P, which is 40 m horizontally from O and 10 m above the level of O. (a) Show that tan²θ- 4tan θ+ 2 = 0. (b) Hence find the two possible values of θ, giving tan θ in exact form. (c) For the smaller value of θ, find the time taken to reach P and state, with a reason, whether the particle is rising or falling at P.

SOLUTION

(a) Horizontally t = 40 ÷(28cos θ). Vertically: 10 = 28sin θ×40/28cos θ - 4.9 ×40²/28²cos²θ = 40tan θ- 10sec²θ. With sec²θ= 1 + tan²θ: 10 = 40tan θ- 10 - 10tan²θ, so tan²θ- 4tan θ+ 2 = 0, as required.

(b) tan θ= (4 ±√8)/2 = 2 ±√2. tan θ = 2 ± √2, θ = 30.4° or 73.7°

(c) θ= 30.36°, horizontal component 28cos θ= 24.160 m s⁻¹, so t = 40 ÷24.160 = 1.656. Then v_y = 28sin θ- 9.8 ×1.656 = -2.07 m s⁻¹ < 0. 1.66 s; falling, because v_y < 0

It is just past its highest point, which is 10.2 m above O.

Take i and j as unit vectors horizontally and vertically upwards. The initial velocity is u = (ucos θ)i + (usin θ)j and the acceleration is -gj. The vector constant-acceleration formulae from M1 give, with the origin at the point of projection,

If the particle starts at position vector r₀, add r₀. The i-component is the horizontal motion and the j-component the vertical motion. "Moving parallel to i" means the j-component of v is zero; "moving at 45° below the horizontal" means the j-component equals minus the i-component.

WORKED EXAMPLE · QUESTION

A particle P is projected from a point O on horizontal ground with velocity (12i + 19.6j) m s⁻¹, where i and j are horizontal and vertically upward unit vectors. P moves freely under gravity. (a) Find the speed of projection and the angle of projection. (b) Find, in terms of t, the position vector and the velocity of P at time t seconds. (c) Find the time at which P is moving at 45° below the horizontal. (d) Find the distance OP at that time.

SOLUTION

(a) Speed = √(12² + 19.6²) = 22.98; tan θ= 19.6 ÷12 gives θ= 58.5°. 23.0 m s⁻¹ at 58.5° above the horizontal

(b) r = 12t i + (19.6t − 4.9t²) j, v = 12 i + (19.6 − 9.8t) j

(c) At 45° below the horizontal the j-component is -12: 19.6 - 9.8t = -12, so t = 31.6 ÷9.8 = 3.2245. It lands when 19.6t - 4.9t² = 0, at t = 4, so it is still in the air. 3.22 s

(d) r = 38.69i + 12.25j, so OP = √(38.69² + 12.25²) = 40.59. 40.6 m

M1.2 Kinematics and M1.1 Vectors

The vertical motion is straight-line motion with constant acceleration, and vector form uses v = u + at and r = ut + ½at².

M2.2 Variable acceleration

Differentiating r = ut - ½gt²j gives v = u - gtj and then a = -gj: projectiles are the constant-acceleration case of calculus-based motion.

M2.4 Work, energy and power

With no air resistance, ½mv² - ½mu² = mgh gives the speed at any height, a fast check on worked example 4. P3 supplies sec²θ= 1 + tan²θ and sin 2θ= 2sin θcos θ.

1. A ball is projected horizontally at 6 m s⁻¹ from a table top 4.9 m above the floor. How long does it take to land, and how far from the table does it land?

2. Explain why the horizontal component of a projectile's velocity stays constant.

3. Find the greatest range on horizontal ground of a particle projected at 21 m s⁻¹.

4. A stone projected at 20° above horizontal ground has range 30 m. At what other angle, with the same speed, would it have the same range?

5. A particle has initial velocity (5i + 7j) m s⁻¹. Find its velocity and speed after 2 s.

1. Vertically: 4.9 = 4.9t², so t = 1 s. Horizontally: 6 ×1 = 6 m. 1 s; 6 m

2. In the model the only force is the weight, which is vertical. With no horizontal force there is no horizontal acceleration, so the horizontal velocity does not change.

3. Maximum range = u²/g at 45°: 21² ÷9.8. 45 m

4. sin 2θ= sin(180° - 2θ), so angles θ and 90° - θ give equal ranges. 70°

5. v = 5i + (7 - 9.8 ×2)j = 5i - 12.6j m s⁻¹; speed = √(25 + 12.6²). 13.6 m s⁻¹