Projectiles and motion with variable velocity Problem set A One question at a time

A tennis ball is hit from ground level. Its velocity at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix} 14 \\ 18 - 9.8t \end{pmatrix}\) m s⁻¹, where the first component is horizontal and the second is upward. It starts at the origin.

1. (a) Find the time at which it reaches its greatest height.

   • At the highest point the vertical velocity is zero.
   • \(18 - 9.8t = 0\)
   • t = 1.84 s

2. (b) Find its displacement from the origin at time \(t\).

   • Integrate each component: \(\mathbf{r} = \int \mathbf{v}\, dt = \begin{pmatrix} 14t + c_1 \\ 18t - 4.9t^2 + c_2 \end{pmatrix}\)
   • \(\mathbf{r} = \mathbf{0}\) when \(t = 0\), so \(c_1 = c_2 = 0\).
   • \(\mathbf{r} = \begin{pmatrix} 14t \\ 18t - 4.9t^2 \end{pmatrix}\)

3. (c) Find its greatest height.

   • Put the time of the top into the vertical displacement.
   • \(h = 18(1.837) - 4.9(1.837)^2\)
   • h = 16.5 m

A golf ball is struck from level ground. Its displacement from the launch point at time \(t\) seconds is \(\mathbf{r} = \begin{pmatrix} 9t \\ 12t - 4.9t^2 \end{pmatrix}\) m, where the first component is horizontal and the second is upward. The ground is level.

1. (a) Find the time at which it lands.

   • It lands when the height is zero.
   • \(12t - 4.9t^2 = 0\)
   • Polynomial solver or graph: \(t = 0\) (the launch) or \(t = 2.45\)
   • Landing time 2.45 s

2. (b) Find how far from the origin it lands.

   • At landing the displacement is \(\mathbf{r} = \begin{pmatrix} 9 \times 2.449 \\ 0 \end{pmatrix} = \begin{pmatrix} 22.0 \\ 0 \end{pmatrix}\)
   • Distance from the origin = 22.0 m

A drone moves in a vertical plane. Its velocity at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix} 2t + 5 \\ 14 - 0.4t^2 \end{pmatrix}\) m s⁻¹. Find, when \(t = 3\):

1. (a) the angle that the velocity makes with the direction of \(\mathbf{i}\)

   • Velocity: \(\mathbf{v} = \begin{pmatrix} 11 \\ 10.4 \end{pmatrix}\)
   • \(\tan\theta = \dfrac{10.4}{11}\)
   • \(\theta = 43.4^\circ\)

2. (b) the speed

   • \(|\mathbf{v}| = \sqrt{11^2 + 10.4^2} = 15.1\)
   • Speed = 15.1 m s⁻¹

3. (c) the magnitude of the acceleration

   • Differentiate: \(\mathbf{a} = \begin{pmatrix} 2 \\ -0.8t \end{pmatrix}\)
   • At \(t = 3\): \(\mathbf{a} = \begin{pmatrix} 2 \\ -2.4 \end{pmatrix}\)
   • \(|\mathbf{a}| = 3.12\) m s⁻²

A skater glides across an ice rink. Its velocity at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix} 8 - 3t \\ 2 + 4t \end{pmatrix}\) m s⁻¹.

1. (a) Find the speed when \(t = 3\).

   • At \(t = 3\): \(\mathbf{v} = \begin{pmatrix} -1 \\ 14 \end{pmatrix}\)
   • \(|\mathbf{v}| = \sqrt{(-1)^2 + 14^2} = 14.0\)
   • Speed = 14.0 m s⁻¹

2. (b) Show that the acceleration is constant and find its magnitude.

   • Differentiate: \(\mathbf{a} = \dfrac{d\mathbf{v}}{dt} = \begin{pmatrix} -3 \\ 4 \end{pmatrix}\)
   • There is no \(t\) in the answer, so the acceleration is constant.
   • \(|\mathbf{a}| = \sqrt{(-3)^2 + 4^2} = 5\) m s⁻²

A stone is thrown from the edge of a pond. Its velocity at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix} 8 \\ 15 - 9.8t \end{pmatrix}\) m s⁻¹, where the first component is horizontal and the second is upward.

1. (a) State the magnitude and direction of the acceleration.

   • Differentiate: \(\mathbf{a} = \begin{pmatrix} 0 \\ -9.8 \end{pmatrix}\)
   • The magnitude is 9.8 m s⁻², and the negative second component means downward.

2. (b) Find the initial speed.

   • At \(t = 0\): \(\mathbf{v} = \begin{pmatrix} 8 \\ 15 \end{pmatrix}\)
   • \(|\mathbf{v}| = \sqrt{8^2 + 15^2} = 17.0\) m s⁻¹

3. (c) Find an expression for the speed at time \(t\), and use technology to find its minimum value.

   • \(s = |\mathbf{v}| = \sqrt{8^2 + (15 - 9.8t)^2}\)
   • Graph \(s\) against \(t\) and use the calculator's minimum tool: \(t = 1.53, s = 8\)
   • Check: the vertical component is zero when \(t = 1.53\), so only the horizontal 8 is left.

A stone is thrown across a frozen lake. Its velocity at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix} 6 \\ 11 - 9.8t \end{pmatrix}\) m s⁻¹, where the first component is horizontal and the second is upward. It starts at the origin on level ground.

1. (a) Find its displacement from the origin at time \(t\).

   • \(\mathbf{r} = \int \mathbf{v}\, dt = \begin{pmatrix} 6t \\ 11t - 4.9t^2 \end{pmatrix}\), with the constants zero because it starts at the origin.

2. (b) Show that it lands approximately 2.24 seconds after it is projected, and find its distance from the origin at that time.

   • \(11t - 4.9t^2 = 0\), so \(t = 0\) or \(t = 2.245\)
   • This is 2.24 s (to 3 s.f.).
   • \(\mathbf{r} = \begin{pmatrix} 6 \times 2.245 \\ 0 \end{pmatrix} = \begin{pmatrix} 13.5 \\ 0 \end{pmatrix}\)
   • Distance from the origin = 13.5 m

3. (c) Show that it lands with the same speed as it was projected with.

   • At \(t = 0\): \(|\mathbf{v}| = \sqrt{6^2 + 11^2} = 12.5\)
   • At landing the vertical component is \(11 - 9.8 \times 2.245 = -11\).
   • \(|\mathbf{v}| = \sqrt{6^2 + (-11)^2} = 12.5\)
   • The speeds are equal.

A hovercraft starts at rest from the origin. Its constant acceleration is \(\mathbf{a} = \begin{pmatrix} 5 \\ -2 \end{pmatrix}\) m s⁻².

1. (a) Find expressions for its velocity and displacement at time \(t\).

   • \(\mathbf{v} = \int \mathbf{a}\, dt = \begin{pmatrix} 5t \\ -2t \end{pmatrix}\) (at rest when \(t = 0\))
   • \(\mathbf{r} = \int \mathbf{v}\, dt = \begin{pmatrix} 2.5t^2 \\ -t^2 \end{pmatrix}\) (at the origin when \(t = 0\))

2. (b) How long does it take for its speed to reach 18 m s⁻¹?

   • \(|\mathbf{v}| = t\sqrt{5^2 + (-2)^2} = 5.385t\)
   • \(5.385t = 18\)
   • \(t = 3.34\) s

3. (c) Plot its position in the \(x\)-\(y\) plane when \(t = 0, 1, 2, 3\). Describe the path.

   • At \(t = 0, 1, 2, 3\) the positions are (0, 0), (2.5, −1), (10, −4), (22.5, −9).
   • Every point has \(y = -0.4x\), so the points lie on one straight line through the origin.
   • The velocity is \(t\) times a fixed vector, so it never changes direction.

A robot moves across a warehouse floor. Its constant acceleration is \(\mathbf{a} = \begin{pmatrix} -3 \\ 2 \end{pmatrix}\) m s⁻². When \(t = 0\) it passes through the origin with velocity \(\begin{pmatrix} 6 \\ 1 \end{pmatrix}\) m s⁻¹.

1. (a) Find the velocity and displacement vectors at time \(t\).

   • \(\mathbf{v} = \int \mathbf{a}\, dt = \begin{pmatrix} -3t + c_1 \\ 2t + c_2 \end{pmatrix}\); when \(t = 0\), \(\mathbf{v} = \begin{pmatrix} 6 \\ 1 \end{pmatrix}\), so \(\mathbf{v} = \begin{pmatrix} 6 - 3t \\ 1 + 2t \end{pmatrix}\)
   • \(\mathbf{r} = \int \mathbf{v}\, dt = \begin{pmatrix} 6t - 1.5t^2 \\ t + t^2 \end{pmatrix}\) (zero constants: it starts at the origin)

2. (b) Find the angle between the initial velocity and the velocity after 2 seconds.

   • At \(t = 2\): \(\mathbf{v} = \begin{pmatrix} 0 \\ 5 \end{pmatrix}\)
   • \(\cos\theta = \dfrac{6 \times 0 + 1 \times 5}{\sqrt{6^2 + 1^2}\,\sqrt{0^2 + 5^2}}\)
   • \(\theta = 80.5^\circ\)

3. (c) Plot the positions at \(t = 0, 1, 2, 3\). Does it move in a straight line?

   • At \(t = 0, 1, 2, 3\) the positions are (0, 0), (4.5, 2), (6, 6), (4.5, 12).
   • The steps between points change direction, so the points do not lie on one line.
   • The path is a curve.

A probe's test ball on a small moon is launched from the origin with initial velocity \(\begin{pmatrix} 10 \\ 7 \end{pmatrix}\) m s⁻¹, where the first component is horizontal and the second is upward. The greatest height reached is 6.6 m. Find, to two significant figures, the gravitational acceleration \(g\) there.

• The vertical velocity is \(7 - gt\), zero at the top when \(t = \dfrac{7}{g}\).
• Vertical displacement: \(7t - \dfrac{1}{2}gt^2\). At the top this is \(\dfrac{7^2}{2g} = 6.6\).
• \(g = \dfrac{7^2}{2 \times 6.6} = 3.712\), so \(g = 3.7\) m s⁻².

A beetle crawls over a tilted board. Its velocity at time \(t\) seconds is \(\mathbf{v} = \begin{pmatrix} 2t - 5 \\ 3 - t \end{pmatrix}\) m s⁻¹ and it starts at the origin. Show that it never returns to its starting position.

• \(\mathbf{r} = \int \mathbf{v}\, dt = \begin{pmatrix} t^2 - 5t \\ 3t - \dfrac{1}{2}t^2 \end{pmatrix}\), with the constants zero.
• It is at the origin only if both components are zero together.
• \(t^2 - 5t = 0\) gives \(t = 0\) or \(t = 5\).
• \(3t - \dfrac{1}{2}t^2 = 0\) gives \(t = 0\) or \(t = 6\).
• For \(t > 0\) the two values are different, so the components are never both zero.