Kinematics and projectile motion: practice worksheet

1. Question 1warm-up · calculator

   A ball is thrown vertically upwards from the ground with an initial speed of $20\ \text{m s}^{-1}$. Air resistance is negligible. Take $g = 10\ \text{m s}^{-2}$ for this question.

   (a)Find the time taken for the ball to reach its maximum height.[1]

   Markscheme

   $v = u + at \Rightarrow 0 = 20 - 10t$ so $t = 2\ \text{s}$ ✔

   1

   Answer: $t = 2\ \text{s}$

   (b)Find the maximum height reached by the ball.[1]

   Markscheme

   $v^{2} = u^{2} + 2as \Rightarrow 0 = 400 - 20s$ so $s = 20\ \text{m}$ ✔

   1

   Accept $s = \frac{(u+v)t}{2} = \frac{20 \times 2}{2} = 20\ \text{m}$ or any other correct equation of motion.

   Answer: $s = 20\ \text{m}$

   (c)State the acceleration of the ball at the instant it is at its maximum height.[1]

   Markscheme

   $10\ \text{m s}^{-2}$ directed downwards / towards the ground ✔

   1

   Both magnitude and direction required. Do not accept zero.

   Answer: $10\ \text{m s}^{-2}$ downwards

   (d)Write down the velocity of the ball when it returns to the point from which it was thrown.[1]

   Markscheme

   $20\ \text{m s}^{-1}$ downwards ✔

   1

   Accept $-20\ \text{m s}^{-1}$ with upwards taken as positive. Direction must be given.

   Answer: $20\ \text{m s}^{-1}$ downwards (i.e. $-20\ \text{m s}^{-1}$ taking upwards as positive)

2. Question 2warm-up · calculator

   A tram moves along a straight track. Its velocity–time graph consists of three straight-line segments: from $t = 0$ to $t = 4\ \text{s}$ the velocity increases uniformly from $0$ to $6\ \text{m s}^{-1}$; from $t = 4\ \text{s}$ to $t = 10\ \text{s}$ the velocity is constant at $6\ \text{m s}^{-1}$; from $t = 10\ \text{s}$ to $t = 12\ \text{s}$ the velocity decreases uniformly to $0$.

   (a)Find the acceleration of the tram during the first $4\ \text{s}$.[1]

   Markscheme

   gradient $= \frac{6 - 0}{4} = 1.5\ \text{m s}^{-2}$ ✔

   1

   Answer: $1.5\ \text{m s}^{-2}$

   (b)Determine the total displacement of the tram between $t = 0$ and $t = 12\ \text{s}$.[2]

   Markscheme

   recognises displacement as area under the graph, e.g. $\frac{1}{2}(4)(6) + (6)(6) + \frac{1}{2}(2)(6)$ ✔	1
   $12 + 36 + 6 = 54\ \text{m}$ ✔	1

   Accept a single trapezium: $\frac{1}{2}(12 + 6)(6) = 54\ \text{m}$.

   Answer: $54\ \text{m}$

   (c)Hence find the average velocity of the tram over the $12\ \text{s}$ journey.[1]

   Markscheme

   $\bar v = \frac{54}{12} = 4.5\ \text{m s}^{-1}$ ✔

   1

   ECF from (b). Do not accept $\frac{0 + 6}{2} = 3\ \text{m s}^{-1}$.

   Answer: $4.5\ \text{m s}^{-1}$

3. Question 3exam-style · calculator

   A stone is thrown horizontally from the top of a vertical sea cliff with a speed of $12.0\ \text{m s}^{-1}$. The top of the cliff is $45.0\ \text{m}$ above the sea. Air resistance is negligible. Take $g = 9.8\ \text{m s}^{-2}$.

   (a)State the horizontal component of the acceleration of the stone while it is in flight.[1]

   Markscheme

   zero ✔

   1

   Answer: zero

   (b)Show that the time taken for the stone to reach the sea is about $3\ \text{s}$.[1]

   Markscheme

   $45.0 = \frac{1}{2}(9.8)t^{2} \Rightarrow t = \sqrt{\frac{2 \times 45.0}{9.8}} = 3.03\ \text{s}$ ✔

   1

   Award the mark for the substituted expression with an unrounded result of $3.03\ \text{s}$.

   Answer: $t = 3.03\ \text{s}$

   (c)Calculate the horizontal distance from the base of the cliff to the point where the stone enters the sea.[1]

   Markscheme

   $x = 12.0 \times 3.03 = 36.4\ \text{m}$ ✔

   1

   Accept $36\ \text{m}$ if $t = 3.0\ \text{s}$ is used.

   Answer: $36.4\ \text{m}$

   (d)Determine the speed of the stone and the angle its velocity makes with the horizontal at the instant it enters the sea.[3]

   Markscheme

   vertical component $v_{y} = 9.8 \times 3.03 = 29.7\ \text{m s}^{-1}$ (or $v_{y} = \sqrt{2 \times 9.8 \times 45.0} = 29.7\ \text{m s}^{-1}$) ✔	1
   $v = \sqrt{12.0^{2} + 29.7^{2}} = 32.0\ \text{m s}^{-1}$ ✔	1
   $\theta = \tan^{-1}\left(\frac{29.7}{12.0}\right) = 68.0^{\circ}$ below the horizontal ✔	1

   Accept $v^{2} = 12.0^{2} + 2(9.8)(45.0)$ giving $32.0\ \text{m s}^{-1}$ directly. Accept $22^{\circ}$ to the vertical. ECF from (b).

   Answer: $v = 32.0\ \text{m s}^{-1}$ at $68.0^{\circ}$ below the horizontal

4. Question 4exam-style · calculator

   A ball is kicked from the flat roof of a building. The ball leaves the roof at a height of $15.0\ \text{m}$ above horizontal ground with a speed of $18.0\ \text{m s}^{-1}$ at an angle of $30^{\circ}$ above the horizontal. Air resistance is negligible. Take $g = 9.8\ \text{m s}^{-2}$.

   (a)Show that the maximum height of the ball above the ground is about $19\ \text{m}$.[2]

   Markscheme

   $u_{y} = 18.0 \sin 30^{\circ} = 9.0\ \text{m s}^{-1}$ ✔	1
   rise above roof $= \frac{9.0^{2}}{2 \times 9.8} = 4.13\ \text{m}$, so height above ground $= 15.0 + 4.13 = 19.1\ \text{m}$ ✔	1

   Award the second mark only if the $15.0\ \text{m}$ is added and an unrounded value is shown.

   Answer: $19.1\ \text{m}$

   (b)Determine the time taken for the ball to reach the ground.[3]

   Markscheme

   uses $s = ut + \frac{1}{2}at^{2}$ with a consistent sign convention, e.g. $-15.0 = 9.0t - 4.9t^{2}$ ✔	1
   solves the quadratic: $t = \frac{9.0 + \sqrt{9.0^{2} + 4(4.9)(15.0)}}{2 \times 4.9}$ ✔	1
   $t = 2.89\ \text{s}$ ✔	1

   ALTERNATIVE: time to top $0.918\ \text{s}$ ✔, fall of $19.1\ \text{m}$ from rest takes $\sqrt{\frac{2 \times 19.1}{9.8}} = 1.98\ \text{s}$ ✔, total $2.89\ \text{s}$ ✔. ECF from (a).

   Answer: $t = 2.89\ \text{s}$

   (c)Calculate the horizontal distance travelled by the ball before it reaches the ground.[1]

   Markscheme

   $x = 18.0 \cos 30^{\circ} \times 2.89 = 45.1\ \text{m}$ ✔

   1

   ECF from (b).

   Answer: $45.1\ \text{m}$

   (d)In reality, air resistance acts on the ball. Explain how air resistance affects the horizontal distance travelled by the ball.[2]

   Markscheme

   air resistance acts opposite to the velocity so there is a horizontal component of force / horizontal deceleration ✔	1
   horizontal component of velocity decreases during flight (is no longer constant) ✔	1
   so the horizontal distance is smaller / the trajectory is asymmetric and steeper on the way down ✔	1

   Award [max 2]. OWTTE.

   Answer: The horizontal distance is reduced because the horizontal component of velocity now decreases during the flight.

5. Question 5exam-style · calculator

   A skydiver of mass $75\ \text{kg}$ steps from a stationary balloon and falls vertically. The table gives the downward velocity $v$ of the skydiver at time $t$ after leaving the balloon.
   
   $$\begin{array}{c|c} t\ /\ \text{s} & v\ /\ \text{m s}^{-1} \\ \hline 0 & 0 \\ 1.0 & 9.7 \\ 2.0 & 18.5 \\ 3.0 & 25.8 \\ 4.0 & 31.6 \\ 5.0 & 35.8 \\ 6.0 & 38.9 \\ 8.0 & 42.3 \\ 10.0 & 43.9 \\ 12.0 & 44.5 \\ 15.0 & 44.9 \\ 20.0 & 45.0 \end{array}$$

   (a)State the terminal speed of the skydiver.[1]

   Markscheme

   $45\ \text{m s}^{-1}$ ✔

   1

   Answer: $45\ \text{m s}^{-1}$

   (b)Estimate the instantaneous acceleration of the skydiver at $t = 4.0\ \text{s}$.[2]

   Markscheme

   acceleration identified as gradient of the velocity–time graph at $t = 4.0\ \text{s}$, e.g. uses $\frac{35.8 - 25.8}{5.0 - 3.0}$ ✔	1
   $a \approx 5.0\ \text{m s}^{-2}$ ✔	1

   Accept values in the range $4.5$ to $5.5\ \text{m s}^{-2}$ from a tangent or symmetric chord about $t = 4.0\ \text{s}$.

   Answer: approximately $5.0\ \text{m s}^{-2}$

   (c)Estimate the distance fallen by the skydiver during the first $6.0\ \text{s}$.[3]

   Markscheme

   distance identified as area under the velocity–time graph ✔	1
   area found by trapezia (or counting strips), e.g. $1.0 \times \left(\frac{0}{2} + 9.7 + 18.5 + 25.8 + 31.6 + 35.8 + \frac{38.9}{2}\right)$ ✔	1
   $s \approx 141\ \text{m}$ ✔	1

   Accept values in the range $135$ to $147\ \text{m}$. Do not award the final mark for $\frac{1}{2}(9.8)(6.0)^{2} = 176\ \text{m}$, which ignores fluid resistance.

   Answer: approximately $140\ \text{m}$

   (d)Explain, with reference to the forces acting on the skydiver, why the acceleration decreases with time and why the velocity eventually becomes constant.[3]

   Markscheme

   at $t = 0$ the only force is the weight so the acceleration is $g$ / $9.8\ \text{m s}^{-2}$ ✔	1
   air resistance (fluid resistance / drag) increases as the speed increases ✔	1
   net force $=$ weight $-$ air resistance decreases, so the acceleration decreases ✔	1
   when air resistance equals the weight the net force is zero, so acceleration is zero and the velocity is constant (terminal speed) ✔	1

   Award [max 3]. OWTTE.

   Answer: The air resistance increases with speed, so the net downward force (weight minus air resistance) decreases and the acceleration decreases; when air resistance equals the weight the net force is zero and the velocity is constant.

6. Question 6challenge · calculator

   A ball of mass $0.150\ \text{kg}$ is thrown from a balcony that is $8.00\ \text{m}$ above horizontal ground. The ball leaves the hand with a speed of $14.0\ \text{m s}^{-1}$ at an angle of $25^{\circ}$ below the horizontal. Air resistance is negligible unless stated otherwise. Take $g = 9.8\ \text{m s}^{-2}$.

   (a)Determine the time taken for the ball to reach the ground.[3]

   Markscheme

   vertical component $u_{y} = 14.0 \sin 25^{\circ} = 5.92\ \text{m s}^{-1}$ downwards ✔	1
   taking downwards as positive: $8.00 = 5.92t + 4.9t^{2}$ ✔	1
   $t = \frac{-5.92 + \sqrt{5.92^{2} + 4(4.9)(8.00)}}{2 \times 4.9} = 0.809\ \text{s}$ ✔	1

   ALTERNATIVE: find $v_{y}$ from $v_{y}^{2} = u_{y}^{2} + 2gs$ giving $13.8\ \text{m s}^{-1}$ ✔, then $t = \frac{v_{y} - u_{y}}{g}$ ✔, $t = 0.809\ \text{s}$ ✔. Accept $0.81\ \text{s}$.

   Answer: $t = 0.809\ \text{s}$

   (b)Show that the speed of the ball just before it hits the ground is about $19\ \text{m s}^{-1}$.[2]

   Markscheme

   $v^{2} = u^{2} + 2gh = 14.0^{2} + 2(9.8)(8.00)$ (or $v_{x} = 12.7\ \text{m s}^{-1}$ and $v_{y} = 13.8\ \text{m s}^{-1}$ combined in quadrature) ✔	1
   $v = \sqrt{352.8} = 18.8\ \text{m s}^{-1}$ ✔	1

   An unrounded value must be shown for the second mark. Accept an energy argument $\frac{1}{2}mv^{2} = \frac{1}{2}mu^{2} + mgh$.

   Answer: $18.8\ \text{m s}^{-1}$

   (c)A second identical ball is thrown from the same balcony with the same speed of $14.0\ \text{m s}^{-1}$, but at $25^{\circ}$ above the horizontal. Without further calculation, state and explain how the impact speed and the time of flight of the second ball compare with those of the first ball.[3]

   Markscheme

   impact speed is the same ✔	1
   because $v^{2} = u^{2} + 2gh$ (or the energy transferred, $mgh$) depends only on the initial speed and the height fallen, not on the launch direction ✔	1
   time of flight is longer because the second ball first rises and returns to the balcony height (with the same downward vertical velocity component $5.92\ \text{m s}^{-1}$) before repeating the motion of the first ball ✔	1

   OWTTE. Do not award the third mark for 'it travels further' without reference to the vertical motion.

   Answer: The impact speed is the same ($18.8\ \text{m s}^{-1}$); the time of flight is longer for the second ball.

   (d)On hitting the ground the first ball sinks into soft mud and comes to rest in $25\ \text{ms}$. Determine the magnitude of the average net force acting on the ball while it is being brought to rest.[3]

   Markscheme

   change in momentum $\Delta p = 0.150 \times 18.8 = 2.82\ \text{kg m s}^{-1}$ ✔	1
   converts time to seconds: $\Delta t = 0.025\ \text{s}$ and uses $F = \frac{\Delta p}{\Delta t}$ ✔	1
   $F = \frac{2.82}{0.025} = 113\ \text{N}$ ✔	1

   Accept $110$ to $115\ \text{N}$ depending on rounding of $v$. ECF from (b). Accept an equivalent approach using $a = \frac{\Delta v}{\Delta t}$ and $F = ma$.

   Answer: $F \approx 113\ \text{N}$

   (e)In reality air resistance acts on the first ball during its flight. Explain how air resistance affects the time of flight and the impact speed of the ball.[2]

   Markscheme

   air resistance acts opposite to the velocity so the (vertical) acceleration is less than $g$ and the ball gains speed more slowly / kinetic energy is transferred to the air ✔	1
   so the impact speed is lower than $18.8\ \text{m s}^{-1}$ ✔	1
   the vertical component of velocity is smaller at every instant, so the ball takes longer to fall $8.00\ \text{m}$ / time of flight increases ✔	1

   Award [max 2]. OWTTE.

   Answer: Both are affected: the impact speed is lower and the time of flight is longer.

Original questions and notes written by Pietro Meloni, PhD, from the IB syllabus. Not endorsed by or affiliated with the International Baccalaureate Organization. Found a mistake? Please tell me privately at pietro@pietromeloni.com. More questions on this topic: A. Space, time and motion practice.