AHL 4.24 Projectiles and constant acceleration · AI HL · AHL 4: Vectors

AHL 4.24

Projectiles and constant acceleration

AHL 4.24

Projectiles and constant acceleration

By the end of this section

Key points

A projectile

A projectile moves in a vertical plane. \(\mathbf{i}\) is horizontal and \(\mathbf{j}\) is upward. Use \(g = 9.8\) m s⁻².
The horizontal velocity stays constant. The vertical velocity falls by 9.8 every second, so \(\mathbf{v} = \begin{pmatrix} p \\ q - 9.8t \end{pmatrix}\) and the acceleration is \(\begin{pmatrix} 0 \\ -9.8 \end{pmatrix}\): magnitude 9.8, directed downward.
Integrate the velocity to get the displacement, \(\mathbf{r} = \int \mathbf{v}\, dt\). The starting position fixes the constants. From the origin they are zero.

Greatest height, landing and speed

Greatest height: set the vertical velocity to zero, find \(t\), then put it into the vertical displacement.
Landing on level ground: set the vertical displacement to zero and solve with technology. \(t = 0\) is the launch. The landing time is twice the time to the top.
At landing the height is zero, so the horizontal component of \(\mathbf{r}\) is the distance from the origin.
A projectile lands with the speed it started with: the horizontal component is unchanged and the vertical component changes sign.
Speed is \(|\mathbf{v}|\). For a projectile it is least when the vertical velocity is zero, and then it equals the horizontal speed.

Velocity, acceleration and angles

Differentiate the velocity to get the acceleration: \(\mathbf{a} = \dfrac{d\mathbf{v}}{dt}\). Substitute \(t\) after differentiating.
The angle of \(\mathbf{v} = \begin{pmatrix} v_x \\ v_y \end{pmatrix}\) with \(\mathbf{i}\) is \(\tan^{-1}\dfrac{v_y}{v_x}\). The angle between two vectors comes from the scalar product.
If \(\mathbf{a}\) has no \(t\) in it, the acceleration is constant.

Constant acceleration

From rest at the origin: \(\mathbf{v} = t\mathbf{a}\) and \(\mathbf{r} = \tfrac{1}{2}t^2\mathbf{a}\). The velocity is \(t\) times a fixed vector, so the direction never changes and the path is a straight line.
With an initial velocity \(\mathbf{u}\): \(\mathbf{v} = \mathbf{u} + t\mathbf{a}\) and \(\mathbf{r} = t\mathbf{u} + \tfrac{1}{2}t^2\mathbf{a}\). The path is usually a curve.
To find when the speed reaches a value, write \(|\mathbf{v}|\) in terms of \(t\) and solve.
To show a particle never returns to the start, find the times when each component of \(\mathbf{r}\) is zero. They never match for \(t > 0\).

On the calculator

Solving for the landing time

Enter the vertical displacement, for example \(25t - 4.9t^2\), as a function of \(t\).
Use the polynomial solver, or graph it and find the zeros.
Ignore \(t = 0\), the launch. The other solution is the landing time.

Screenshot and video to come

Minimum speed from a graph

Enter the speed \(s(t) = \sqrt{p^2 + (q - 9.8t)^2}\) as a function.
Graph it with a window that shows the lowest point.
Use the minimum tool. The x-value is the time and the y-value is the least speed.

Screenshot and video to come

Explore: a projectile under gravity

The velocity is (p, q − 9.8t). Drag the tip of the orange launch arrow to set p and q, and drag the launch point up to give a height. Move the time slider to follow the object. Before you press "Show the landing point", type your prediction of the range.

Horizontal velocity, p (m s⁻¹) Vertical velocity at the start, q (m s⁻¹)

Launch height, h (m) Time, t (s)

A wall (drag its top)

i is horizontal and j is upward. The acceleration is (0, −9.8). The black dot is the object. The orange arrow is the launch velocity and the teal arrow the velocity now, both to the same scale; the dashed lines are the components.

My prediction of the range m

Move the time slider to follow the object.

Set the launch height to 0. How does the time of flight compare with the time to the top? Why?
Move the time slider. Which part of the velocity changes and which part stays the same? What in the acceleration explains this?
Raise the launch point. Is the landing time still twice the time to the top? Is the landing speed still the launch speed?
Choose "Speed and angle" with θ = 45° and add the wall. Change u until the object only just clears the wall. What is the least speed? Check it by solving for u.

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.

Highest point and landing of a ball

A ball is thrown from the origin on level ground. Its velocity at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix} 12 \\ 20 - 9.8t \end{pmatrix}\) m s⁻¹, where the first component is horizontal and the second is upward. (a) Find the time at which the ball reaches its greatest height. (b) Find its displacement from the origin at time \(t\), and hence its greatest height. (c) How far from the origin does the ball land?

Working

Time to the top
\(20 - 9.8t = 0\), \(t = 2.04\) s

At the highest point the vertical velocity is zero.
Displacement
\(\mathbf{r} = \int \mathbf{v}\, dt = \begin{pmatrix} 12t + c_1 \\ 20t - 4.9t^2 + c_2 \end{pmatrix} = \begin{pmatrix} 12t \\ 20t - 4.9t^2 \end{pmatrix}\)

The ball starts at the origin, so both constants are zero.
Greatest height
\(h = 20(2.041) - 4.9(2.041)^2 = 20.4\) m

Put the time of the top into the vertical displacement.
Landing time
Polynomial solver: \(20t - 4.9t^2 = 0\), \(t = 4.08\)

The other solution, \(t = 0\), is the throw. The landing time is twice the time to the top: \(2 \times 2.04\).
Distance landed
\(\mathbf{r} = \begin{pmatrix} 12 \times 4.082 \\ 0 \end{pmatrix} = \begin{pmatrix} 49.0 \\ 0 \end{pmatrix}\), so 49.0 m

The height is zero, so the distance from the origin is the horizontal component.

Angle, speed and acceleration from a velocity

An object moves with velocity \(\mathbf{v} = \begin{pmatrix} 4t + 1 \\ 15 - 0.3t^2 \end{pmatrix}\) m s⁻¹ at time \(t\) seconds. When \(t = 5\), find (a) the angle that the velocity makes with the direction of \(\mathbf{i}\), (b) the speed, (c) the magnitude of the acceleration.

Working

Velocity at t = 5
\(\mathbf{v} = \begin{pmatrix} 4(5) + 1 \\ 15 - 0.3(25) \end{pmatrix} = \begin{pmatrix} 21 \\ 7.5 \end{pmatrix}\)

Substitute \(t\) into each component.
Angle with i
\(\tan\theta = \dfrac{7.5}{21}\), \(\theta = 19.7^\circ\)

The horizontal component is along \(\mathbf{i}\).
Speed
\(|\mathbf{v}| = \sqrt{21^2 + 7.5^2} = 22.3\) m s⁻¹

Speed is the magnitude of the velocity.
Acceleration
\(\mathbf{a} = \dfrac{d\mathbf{v}}{dt} = \begin{pmatrix} 4 \\ -0.6t \end{pmatrix} = \begin{pmatrix} 4 \\ -3 \end{pmatrix}\), \(|\mathbf{a}| = 5\) m s⁻²

Differentiate each component first, then put in \(t = 5\).

Acceleration and minimum speed of a projectile

The velocity of a projectile is \(\mathbf{v} = \begin{pmatrix} 10 \\ 15 - 9.8t \end{pmatrix}\) m s⁻¹, where the first component is horizontal and the second is upward. (a) State the magnitude and direction of its acceleration. (b) Find its initial speed. (c) Find an expression for its speed at time \(t\), and use technology to find the minimum speed.

Working

Acceleration
\(\mathbf{a} = \begin{pmatrix} 0 \\ -9.8 \end{pmatrix}\)

Differentiate. The magnitude is 9.8 m s⁻² and it points downward.
Initial speed
\(|\mathbf{v}(0)| = \sqrt{10^2 + 15^2} = 18.0\) m s⁻¹

Put \(t = 0\) into the velocity.
Speed at time t
\(s = \sqrt{10^2 + (15 - 9.8t)^2}\)

Speed is the magnitude of the velocity.
Minimum
From the calculator: \(t = 1.53, s = 10\)

Graph \(s\) and find the minimum. The vertical component is zero at that moment.

Landing with the speed it started with

A ball is kicked from the origin on level ground with velocity \(\mathbf{v} = \begin{pmatrix} 8 \\ 6 - 9.8t \end{pmatrix}\) m s⁻¹, where the first component is horizontal and the second is upward. Find the time at which it lands, and show that it lands with the same speed as it was kicked with.

Working

Displacement
\(\mathbf{r} = \begin{pmatrix} 8t \\ 6t - 4.9t^2 \end{pmatrix}\)

Integrate, with zero constants for a start at the origin.
Landing time
\(6t - 4.9t^2 = 0\), \(t = 1.22\) s

Technology gives \(t = 0\) or the landing time.
Velocity at landing
\(\mathbf{v} = \begin{pmatrix} 8 \\ 6 - 9.8 \times 1.224 \end{pmatrix} = \begin{pmatrix} 8 \\ -6 \end{pmatrix}\)

The vertical velocity has changed sign.
Compare
\(\sqrt{8^2 + 6^2} = 10\) at the start, \(\sqrt{8^2 + (-6)^2} = 10\) at landing

Same horizontal component, opposite vertical component: the same speed.

Constant acceleration from rest

An object starts at rest at the origin and has constant acceleration \(\mathbf{a} = \begin{pmatrix} 2 \\ -5 \end{pmatrix}\) m s⁻². (a) Find its velocity and displacement at time \(t\). (b) How long does it take for its speed to reach 15 m s⁻¹? (c) Plot its position when \(t = 0, 1, 2, 3\) and describe the path.

Working

Velocity
\(\mathbf{v} = \int \mathbf{a}\, dt = \begin{pmatrix} 2t \\ -5t \end{pmatrix}\)

At rest when \(t = 0\), so the constants are zero.
Displacement
\(\mathbf{r} = \int \mathbf{v}\, dt = \begin{pmatrix} t^2 \\ -2.5t^2 \end{pmatrix}\)

At the origin when \(t = 0\).
Time for speed 15
\(|\mathbf{v}| = t\sqrt{2^2 + 5^2} = 15\), \(t = \dfrac{15}{\sqrt{29}} = 2.79\) s

The speed is the magnitude of the velocity.
Positions
(0, 0), (1, −2.5), (4, −10), (9, −22.5)

Put \(t = 0, 1, 2, 3\) into the displacement.
The path
A straight line, \(y = -2.5x\)

The velocity is \(t\) times a fixed vector, so its direction never changes.

Constant acceleration with an initial velocity

An object has constant acceleration \(\mathbf{a} = \begin{pmatrix} -1 \\ 2 \end{pmatrix}\) m s⁻². When \(t = 0\) it passes through the origin with velocity \(\begin{pmatrix} 3 \\ 1 \end{pmatrix}\) m s⁻¹. (a) Find its velocity and displacement at time \(t\). (b) Find the angle between its initial velocity and its velocity after 2 seconds. (c) Plot its position when \(t = 0, 1, 2, 3\). Does it move in a straight line?

Working

Velocity
\(\mathbf{v} = \begin{pmatrix} 3 - t \\ 1 + 2t \end{pmatrix}\)

Integrate the acceleration and use the initial velocity to fix the constants.
Displacement
\(\mathbf{r} = \begin{pmatrix} 3t - 0.5t^2 \\ t + t^2 \end{pmatrix}\)

Integrate again. It starts at the origin.
Velocity at t = 2
\(\mathbf{v} = \begin{pmatrix} 1 \\ 5 \end{pmatrix}\)

Put \(t = 2\) into the velocity.
The angle
\(\cos\theta = \dfrac{3 \times 1 + 1 \times 5}{\sqrt{10}\,\sqrt{26}}\), \(\theta = 60.3^\circ\)

Use the scalar product of the two velocity vectors.
The path
(0, 0), (2.5, 2), (4, 6), (4.5, 12): not in a line

The steps between the points change direction, so the path is a curve.

Finding g from the greatest height

A ball is thrown from the origin on another planet with initial velocity \(\begin{pmatrix} 5 \\ 8 \end{pmatrix}\) m s⁻¹, where the first component is horizontal and the second is upward. The greatest height it reaches is 6.1 m. Find, to two significant figures, the gravitational acceleration \(g\) on that planet.

Working

Vertical velocity
\(8 - gt = 0\), \(t = \dfrac{8}{g}\)

At the top the vertical velocity is zero.
Height at the top
\(8t - \dfrac{1}{2}gt^2 = \dfrac{8^2}{2g}\)

Put \(t = 8/g\) into the vertical displacement.
Solve for g
\(\dfrac{64}{2g} = 6.1\), \(g = 5.246\), so \(g = 5.2\) m s⁻²

Set the height equal to 6.1.

A particle that never returns

The velocity of a particle at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix} 4t - 6 \\ 3 - t \end{pmatrix}\) m s⁻¹, and it starts at the origin. Show that it never returns to its starting position.

Working

Displacement
\(\mathbf{r} = \begin{pmatrix} 2t^2 - 6t \\ 3t - 0.5t^2 \end{pmatrix}\)

Integrate, with zero constants.
x-component zero
\(2t(t - 3) = 0\), so \(t = 0\) or \(t = 3\)

The particle is back at \(x = 0\) only at these times.
y-component zero
\(t(3 - 0.5t) = 0\), so \(t = 0\) or \(t = 6\)

The particle is back at \(y = 0\) only at these times.
Conclusion
For \(t > 0\) the times 3 and 6 are different.

Both components must be zero at the same time to return to the origin. They never are.

Menu

Vectors

Projectiles and constant acceleration

By the end of this section

Key points

A projectile

Greatest height, landing and speed

Velocity, acceleration and angles

Constant acceleration

On the calculator

Solving for the landing time

Minimum speed from a graph

Explore: a projectile under gravity

Worked examples

Highest point and landing of a ball

Working

Angle, speed and acceleration from a velocity

Working

Acceleration and minimum speed of a projectile

Working

Landing with the speed it started with

Working

Constant acceleration from rest

Working

Constant acceleration with an initial velocity

Working

Finding g from the greatest height

Working

A particle that never returns

Working

Choose a skill

Quiz