Vertical Motion Under Gravity

AS · M1 · 13 min

A ball thrown straight up, a stone dropped down a well, a sandbag released from a rising balloon: all move in a vertical line with the same constant acceleration, the acceleration due to gravity. That makes them direct applications of the constant acceleration formulae, with one extra demand: you must handle signs carefully, because the particle often moves up and then down within a single stage. Questions on vertical motion appear regularly in Paper 4, sometimes on their own and often as the setting for an energy question.

The model

In Paper 4, a particle moving vertically under gravity is modelled with these assumptions:

the object is a particle (its size and any rotation are ignored);
there is no air resistance;
the acceleration due to gravity is constant, $g = 10\ \text{m s}^{-2}$ , directed vertically downwards.

Key result

A particle moving freely under gravity has constant acceleration $g = 10\ \text{m s}^{-2}$ downwards, whichever way it is moving, including at the instant it is at its highest point.

Cambridge specifies $g = 10\ \text{m s}^{-2}$ for 9709 Paper 4. Using $9.8$ or $9.81$ gives different answers and can cost accuracy marks.

The acceleration does not depend on the mass. A heavy stone and a light stone dropped together (with no air resistance) fall side by side. This is why masses rarely appear in pure kinematics questions on vertical motion.

Signs: choosing up or down as positive

Choose a positive direction and stick with it for the whole stage.

Upwards positive is the usual choice when the particle is projected upwards. Then:

$a = -10$ throughout;
$v > 0$ while the particle rises, $v = 0$ at the top, $v < 0$ while it falls;
$s$ is the displacement above the point of projection; it is negative when the particle is below that point.

Downwards positive is convenient for something dropped or thrown downwards, where the motion is all in one direction. Then $a = +10$ and all the quantities are positive.

Watch out

The most common error in this topic is mixing signs: taking $u$ upwards as positive but using $a = +10$ , or measuring $s$ downwards to the ground while $u$ is measured upwards. Write "taking upwards as positive" at the top of your working and give every value its sign accordingly.

Projected vertically upwards

Consider a ball projected vertically upwards from ground level with speed $u$ . Taking upwards as positive, $a = -10$ .

Time to the highest point. At the top, $v = 0$ :

0 = u - 10t \quad\Rightarrow\quad t = \frac{u}{10}

Greatest height. Using $v^2 = u^2 + 2as$ with $v = 0$ :

0 = u^2 - 20s \quad\Rightarrow\quad s = \frac{u^2}{20}

Return to the starting level. When the ball is back at the start, $s = 0$ :

0 = ut - 5t^2 = t(u - 5t) \quad\Rightarrow\quad t = 0 \text{ or } t = \frac{2u}{10}

So the time to come back down equals the time to go up. The velocity on return is $u - 10 \times \tfrac{2u}{10} = -u$ : the same speed, now downwards.

Projected upwards with speed u (no air resistance)

Time to greatest height $= \dfrac{u}{g}$ ; greatest height $= \dfrac{u^2}{2g}$ .
Time to return to the point of projection $= \dfrac{2u}{g}$ .
At any given height the speed going up equals the speed coming down.

The motion is symmetrical about the highest point. This symmetry is a useful check, but in an exam it is safer to derive results with suvat than to quote them.

The velocity–time graph is a single straight line with gradient $-10$ . The ball below is projected at $15\ \text{m s}^{-1}$ . The graph crosses the axis at $t = 1.5$ (the highest point), and the two triangles have equal areas, so the displacement after $3\ \text{s}$ is zero: the ball is back where it started.

y = 15 - 10x (3, 0) -- (3, -15) fill 0 1.5 y = 15 - 10x fill 1.5 3 y = 15 - 10x

The displacement–time graph is a parabola, $s = 15t - 5t^2$ , with its vertex at the greatest height $11.25\ \text{m}$ when $t = 1.5$ .

y = 15x - 5x^2 (1.5, 0) -- (1.5, 11.25)

Greatest height and time of flight

A ball is projected vertically upwards from ground level with speed $15\ \text{m s}^{-1}$ . Find (a) the greatest height reached, (b) the time taken to reach this height, (c) the total time before the ball returns to the ground.

Solution

Take upwards as positive: $u = 15$ , $a = -10$ .

(a) At the greatest height $v = 0$ . Using $v^2 = u^2 + 2as$ :

0 = 15^2 - 20s \quad\Rightarrow\quad s = \frac{225}{20} = 11.25\ \text{m}

(b) Using $v = u + at$ : $0 = 15 - 10t$ , so $t = 1.5\ \text{s}$ .

Starting above the ground

If the particle starts above the ground (from a cliff, a tower or a window), it lands below its starting point. With upwards positive, its displacement on landing is negative. Working the whole flight as one stage is quicker and safer than splitting it at the top.

A stone projected vertically upwards from the edge of a cliff. It rises, stops, then falls past the cliff top into the sea. With upwards positive, it hits the sea when s = −20.

Projected upwards from a cliff

A stone is projected vertically upwards with speed $8\ \text{m s}^{-1}$ from a point at the edge of a cliff, $20\ \text{m}$ above the sea. Find the time taken for the stone to reach the sea and its speed as it hits the water.

Solution

Take upwards as positive. The sea is $20\ \text{m}$ below the point of projection: $s = -20$ , $u = 8$ , $a = -10$ .

-20 = 8t - 5t^2 \quad\Rightarrow\quad 5t^2 - 8t - 20 = 0

t = \frac{8 \pm \sqrt{64 + 400}}{10} = \frac{8 \pm 21.54}{10}

The negative root has no meaning here, so $t = 2.95\ \text{s}$ .

For the speed, $v^2 = u^2 + 2as = 64 + 2(-10)(-20) = 464$ , so $v = -21.5$ (negative, because the stone is moving down). The speed is $21.5\ \text{m s}^{-1}$ .

Note that $v^2 = u^2 + 2as$ gives the speed without needing the time, so it does not inherit any rounding from the quadratic.

Objects released from moving things

Something dropped from a moving object starts with the velocity of that object, not from rest. A sandbag let go from a balloon rising at $5\ \text{m s}^{-1}$ first moves upwards at $5\ \text{m s}^{-1}$ , slows, stops and then falls.

A sandbag released from a rising balloon

A balloon is rising vertically with constant speed $5\ \text{m s}^{-1}$ . When it is $30\ \text{m}$ above the ground, a small sandbag is released from it. Find the time taken for the sandbag to reach the ground and its speed when it lands.

Solution

At release the sandbag is moving upwards with the balloon. Taking upwards as positive: $u = 5$ , $a = -10$ , $s = -30$ .

-30 = 5t - 5t^2 \quad\Rightarrow\quad t^2 - t - 6 = 0 \quad\Rightarrow\quad (t - 3)(t + 2) = 0

So $t = 3\ \text{s}$ . Then $v = 5 - 10(3) = -25$ : the sandbag lands at $25\ \text{m s}^{-1}$ .

Taking $u = 0$ here would give $t = 2.45\ \text{s}$ : wrong, because it ignores the sandbag's upward velocity at release.

Time spent above a given height

A particle projected upwards passes a given height twice, once rising and once falling. Setting $s$ equal to that height gives a quadratic in $t$ whose two roots are those two times. The time spent above the height is the difference between them.

Time above a height

A ball is projected vertically upwards from a point $O$ with speed $20\ \text{m s}^{-1}$ . Find the length of time for which the ball is at least $15\ \text{m}$ above $O$ .

Solution

Upwards positive: $s = 15$ , $u = 20$ , $a = -10$ .

15 = 20t - 5t^2 \quad\Rightarrow\quad t^2 - 4t + 3 = 0 \quad\Rightarrow\quad (t - 1)(t - 3) = 0

The ball is $15\ \text{m}$ above $O$ at $t = 1$ (rising) and $t = 3$ (falling), so it is at least $15\ \text{m}$ above $O$ for $3 - 1 = 2\ \text{s}$ .

Two particles moving vertically

Two-particle questions work exactly as in Equations of constant acceleration: write the height of each above a common level in terms of a common time, then set them equal (or set their difference to a given value). If one particle is projected $T$ seconds after the other, its time is $t - T$ .

A neat fact helps with checking: because both particles have the same acceleration, the $-5t^2$ terms usually cancel, leaving a linear equation.

One dropped, one thrown up

A particle $A$ is projected vertically upwards from ground level with speed $25\ \text{m s}^{-1}$ . At the same instant, a particle $B$ is released from rest at a point $50\ \text{m}$ vertically above $A$ 's starting point. Find the time at which the particles collide, the height at which this happens, and the velocity of each just before the collision.

Solution

Take upwards as positive and measure heights from the ground.

Height of $A$ : $h_A = 25t - 5t^2$ . Height of $B$ : $h_B = 50 - 5t^2$ .

They collide when $h_A = h_B$ :

25t - 5t^2 = 50 - 5t^2 \quad\Rightarrow\quad 25t = 50 \quad\Rightarrow\quad t = 2\ \text{s}

Height $= 50 - 5(4) = 30\ \text{m}$ above the ground.

Velocities: $A$ has $v = 25 - 10(2) = 5\ \text{m s}^{-1}$ (upwards); $B$ has $v = 0 - 10(2) = -20$ , that is $20\ \text{m s}^{-1}$ downwards.

Two particles projected from the same point (exam standard)

A particle $P$ is projected vertically upwards from a point $O$ with speed $20\ \text{m s}^{-1}$ . One second later a second particle $Q$ is projected vertically upwards from $O$ with speed $25\ \text{m s}^{-1}$ . Find the time after $P$ is projected at which the two particles are at the same height, find this height, and describe the motion of each particle at that instant.

Solution

Take upwards as positive, with $t$ the time after $P$ is projected.

h_P = 20t - 5t^2, \qquad h_Q = 25(t - 1) - 5(t - 1)^2 \quad (t \ge 1)

Expand $h_Q = 25t - 25 - 5t^2 + 10t - 5 = 35t - 30 - 5t^2$ . Setting $h_P = h_Q$ :

20t - 5t^2 = 35t - 30 - 5t^2 \quad\Rightarrow\quad 15t = 30 \quad\Rightarrow\quad t = 2\ \text{s}

Height $= 20(2) - 5(4) = 20\ \text{m}$ above $O$ .

Velocities at $t = 2$ : $P$ has $v = 20 - 10(2) = 0$ , so $P$ is momentarily at rest at its highest point ( $\tfrac{20^2}{20} = 20\ \text{m}$ , which agrees). $Q$ has $v = 25 - 10(1) = 15\ \text{m s}^{-1}$ upwards.

Check: $h_Q = 25(1) - 5(1) = 20$ . Correct.

Distance versus displacement in vertical motion

For a ball that goes up and comes down, the displacement can be small while the distance travelled is large. To find a distance, split the motion at the highest point, as in the constant acceleration note: find the greatest height, then add the distance fallen.

Common mistakes

Vertical motion errors

Acceleration zero at the top. At the highest point the velocity is zero but the acceleration is still $10\ \text{m s}^{-2}$ downwards.
Inconsistent signs. If $u$ is positive upwards then $a = -10$ and a point below the start has negative $s$ .
Taking $u = 0$ for an object released from a moving balloon or lift. It starts with the velocity of whatever released it.
Splitting the flight at the top when you do not need to. One stage with $s$ negative handles a stone thrown up from a cliff in one equation.
Using $g = 9.8$ or $9.81$ . The 9709 syllabus says use $g = 10\ \text{m s}^{-2}$ .
Giving a negative speed. Velocity can be negative; speed is its size.

Exam technique

Exam tip

State "taking upwards as positive" (or downwards) at the start. Examiners award the equation mark only if the signs are consistent.
When a quadratic gives two times, say which one you are using and why ("the negative root is rejected", or "the ball passes this height at $t = 1$ on the way up and $t = 3$ on the way down").
"Find the speed with which it hits the ground" is often quicker by $v^2 = u^2 + 2as$ than by first finding the time.
Model-based questions sometimes ask for an assumption: the standard answers are "no air resistance" and "the object can be modelled as a particle".
Vertical motion questions are often the first part of an energy question; the same answer can be found by conservation of energy, which makes a good check.

Summary

Free vertical motion: constant acceleration $g = 10\ \text{m s}^{-2}$ downwards, independent of mass, no air resistance.
Choose up or down as positive and keep it; with up positive, $a = -10$ throughout.
At the highest point $v = 0$ but $a = -10$ .
Projected upwards with speed $u$ : greatest height $\tfrac{u^2}{20}$ , time to top $\tfrac{u}{10}$ , back at the start after $\tfrac{2u}{10}$ with speed $u$ .
A particle below its starting point has negative displacement (with up positive): use one stage for the whole flight.
Objects released from a moving body start with that body's velocity.
Time above a height $=$ difference between the two roots of the quadratic in $t$ .

Practice

Question

A stone is dropped from rest from the top of a tower $45\ \text{m}$ high. Find the time it takes to reach the ground and its speed on impact.
A ball is thrown vertically upwards with speed $12\ \text{m s}^{-1}$ . Find the greatest height above the point of projection and the time taken to return to the point of projection.
A ball is thrown vertically downwards with speed $4\ \text{m s}^{-1}$ from a window $25\ \text{m}$ above the ground. Find the time taken to reach the ground and the speed on impact.
A particle projected vertically upwards from the ground reaches a greatest height of $20\ \text{m}$ . Find its speed of projection and the total time it is in the air.
A ball is projected vertically upwards from ground level with speed $18\ \text{m s}^{-1}$ . Find the length of time for which it is at least $9\ \text{m}$ above the ground.
A particle is projected vertically upwards from a point $15\ \text{m}$ above the ground and reaches the ground $3\ \text{s}$ later. Find the speed of projection and the greatest height of the particle above the ground.
A stone is dropped from rest from the top of a cliff $80\ \text{m}$ above the sea. One second later a second stone is thrown vertically downwards from the same point. Both stones hit the sea at the same instant. Find the speed with which the second stone was thrown.
A particle $A$ is projected vertically upwards from a point $O$ with speed $30\ \text{m s}^{-1}$ . Two seconds later a particle $B$ is projected vertically upwards from $O$ , also with speed $30\ \text{m s}^{-1}$ . Find the time after $A$ was projected at which the particles collide, the height above $O$ at which they collide, and the velocity of each just before the collision.
A ball is projected vertically upwards from ground level with speed $14\ \text{m s}^{-1}$ . Find the distance travelled by the ball in the first $2\ \text{s}$ of its motion and its average speed over that time.

Answers

Downwards positive: $45 = 5t^2$ , so $t = 3\ \text{s}$ ; $v = 10(3) = 30\ \text{m s}^{-1}$ .
Upwards positive: $0 = 144 - 20s$ , so $s = 7.2\ \text{m}$ . $0 = 12t - 5t^2$ gives $t = 2.4\ \text{s}$ .
Downwards positive: $25 = 4t + 5t^2$ , so $5t^2 + 4t - 25 = 0$ and $t = \dfrac{-4 + \sqrt{516}}{10} = 1.87\ \text{s}$ . $v^2 = 16 + 2(10)(25) = 516$ , so $v = 22.7\ \text{m s}^{-1}$ .
$0 = u^2 - 20(20)$ , so $u = 20\ \text{m s}^{-1}$ . Back to the ground: $0 = 20t - 5t^2$ , so $t = 4\ \text{s}$ .
$9 = 18t - 5t^2$ gives $5t^2 - 18t + 9 = 0$ , so $(5t - 3)(t - 3) = 0$ and $t = 0.6$ or $3$ . Time above $9\ \text{m}$ : $3 - 0.6 = 2.4\ \text{s}$ .
Upwards positive, $s = -15$ : $-15 = 3u - 5(9)$ , so $u = 10\ \text{m s}^{-1}$ . Greatest height above the point of projection $= \tfrac{100}{20} = 5\ \text{m}$ , so $20\ \text{m}$ above the ground.
First stone: $80 = 5t^2$ , so it takes $4\ \text{s}$ . The second stone has $3\ \text{s}$ : $80 = 3u + 5(9)$ , so $u = \tfrac{35}{3} = 11.7\ \text{m s}^{-1}$ .
$30t - 5t^2 = 30(t - 2) - 5(t - 2)^2$ . Expanding the right side: $30t - 60 - 5t^2 + 20t - 20$ . So $0 = 20t - 80$ and $t = 4\ \text{s}$ . Height $= 120 - 80 = 40\ \text{m}$ . $A$ : $v = 30 - 40 = -10$ , so $10\ \text{m s}^{-1}$ downwards; $B$ : $v = 30 - 20 = 10\ \text{m s}^{-1}$ upwards.
Top at $t = 1.4\ \text{s}$ , height $\tfrac{196}{20} = 9.8\ \text{m}$ . At $t = 2$ : $s = 28 - 20 = 8\ \text{m}$ . Distance $= 9.8 + (9.8 - 8) = 11.6\ \text{m}$ . Average speed $= \tfrac{11.6}{2} = 5.8\ \text{m s}^{-1}$ .

The model

Signs: choosing up or down as positive

Projected vertically upwards

Starting above the ground

Objects released from moving things

Time spent above a given height

Two particles moving vertically

Distance versus displacement in vertical motion

Common mistakes

Exam technique

Summary

Practice

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