Monday, 20 September 2021 12:21

VERTICAL PROJECTILE MOTION IN ONE DIMENSION QUESTIONS AND ANSWERS

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Activity 1
A ball of mass 200 g is dropped from the roof of a 100 m high building.
Ignore air resistance and calculate:

  1. the velocity of the ball when it hits the ground. (4)
  2. how long the ball is in the air before it hits the ground. (5)
    [9]

Solutions
(Calculations for ‘down positive’ and for ‘up positive’ are provided. You only need to do one way!)
Let direction of motion down be positive
vi = 0 m·s−1            vf = ? (a)
Δy = + 100 m         a = + 9,8 m·s−2
Δt = ? (b)

  1. v2f = v2i + 2a · Δy
    = 02 + (2)(9,8)(100)
    = 1960
    vf = √1960 = +44,27 m·s–1
    ∴ 44,27 m·s–1, downwards (4)
  2. vf = vi + a Δt
    44,27 = 0 + (9,8) Δt
    Δt = 44,27 = 4,52 s
             9.8
    ∴ the ball is in the air for 4,52 s (5)
    [9] 

If we let direction of motion up to be positive the solution is the same, only the sign changes.
This example shows that projectiles can have their motion described by a single set of equations for both upward and downward motion. It is not necessary to set motion in two directions for the same question.
However it is important that you are able to solve problems using both approaches i.e. downwards - positive OR upwards - positive.

Activity 2
A ball is thrown vertically upwards at 4 m·s−1 and returns to the thrower’s hand.
Let the direction of motion downwards be positive.
Calculate:

  1. The maximum height reached by the ball. (4)
  2. The time taken for the ball to reach the highest point in its trajectory. (5)
  3. The total time that the ball is in the air. (3)
  4. The ball’s total displacement during the motion. (1)
    [13]

Solutions
Let the direction of motion down be positive
vi = −4 m·s−1
vf = 0 m·s−1
Δy = ? (a) m
a = + 9,8 m·s−2
Δt = ? (b)

  1. v2f = v2i + 2a · Δy
    0 = (–4)2 + (2)(9,8) yΔ
    19,6 Δy = –16
    Δy = –0,82 m
    ∴ the ball reaches a height of 0,82 m above the starting level (4)
  2. vf =vi + a Δt
    0 = (–4) + (9,8)Δt
    Δt = 4  = 0,41 s
          9.8
    ∴ the ball takes 0,41 s to reach the highest point in its trajectory (5)
  3. Time upwards = time downwards
    ∴ total time in the air is (2)(0,41) = 0,82 s (3)
  4. Total displacement = Δy = 0m (1)
    Displacement is measured in a straight line from the initial position (the thrower’s line from the original to the final position (the thrower’s hand is the initial and final position).
    [13] 

Activity 3
Lerato throws a stone vertically into the air from the top of a cliff. The stone strikes the ground below after 3 s. The velocity vs. time graph below shows the motion of the stone. Ignore the effect of air resistance.
14

Hint:
Use a ruler to draw the axes and any straight lines!
Drawing a sketch graph

  • The graph does not have to be to scale, but it must have the correct shape
  • The graph must show the physical quantity in words followed by the abbreviated unit (in brackets) e.g. velocity (m·s−1)
  • The graph must include the values asked for in the question.
  1. How long does the stone take to fall from the height of the cliff to the ground below? (2)
  2. What is the maximum height that the stone reaches above the groud? (Hint: calculate the height the stone reaches above the cliff, then calculate the height of the cliff, and add these two numbers). (4)
  3. Draw a graph of position versus time. Use upwards as negative. (6)
    [12]

Solutions

  1. 3 - 2,043 = 0,96 s (2)
  2. Option 1
    Δy = vi Δt + ½ a Δt2 
    = (10)(3) + ½ (–9,8)(3)2 
    = 14,1 m
    Δy = 14,1 m below the starting point
    vf2 = vi2 + 2aΔy
    0 = 100 + 2(–9,8) Δy 
    Δy = 5,1 m
    Maximum height above the ground = 5,1 + 14,1 = 19,2m 
    Option 2
    Δy = vi Δt + ½ a Δt2 
    = 0 3 + ½ (–9,8)(3 – 1,02)2 
    = –19,21 m
    Δy = 19,21 m (maximum height above the ground) 
    Option 3
    vf2 = vi
    2 + 2aΔy 
    (–19,4)2 = 0 + 2 (–9,8) Δy 
    Δy = 19,2 m (maximum height above the ground) (4)

  3. 15
    Marks for: correct shape; graph starts at zero; maximum height shown as –5,1 m ; times indicated correctly; graph ends at 3 s . (6)
    [12] 

Activity 4
A hot-air balloon is rising upwards at a constant velocity of 5 ms–1. When the balloon is 60 m above the ground, a boy drops a ball from it and the ball falls freely.
Assume that the balloon continues to move upwards at the same constant velocity. When the ball hits the ground, it bounces vertically upwards to a height of 8 m above the ground. It falls back to the ground and bounces again to reach a height of 5 m. Take upwards as positive.

  1. Describe the motion of the ball from the moment it is dropped until it hits the ground. (8)
  2. Why does the ball not reach the same height during the second bounce as during the first? (7)
  3. What is the magnitude and direction of the ball’s velocity at the moment when it is dropped? (1)
  4. Calculate maximum height reached by the ball. (5)
  5. Where is the ball after 3 seconds? (6)
  6. How far apart will the ball and the balloon be after 3 seconds? (7)
  7. Calculate the time taken for the ball to reach the ground. (7)
  8. Calculate the time the ball takes to reach the height of 8 m above the ground after its first bounce. (10)
  9. Calculate the velocity at which the ball hits the ground the after the first bounce. (4)
  10. Draw a sketch graph of velocity vs. time for the ball from the moment it is dropped until it reaches the height of 5 m after its first bounce. (5)
  11. Draw a sketch graph of position vs. time for the ball for the same time as in (10). Use the position of the ball when it is dropped as the point of reference. (5)
  12. Draw a sketch graph of acceleration vs time for ball for the same time as in (10). (4)
    [69]

Solutions

  1. Initially the ball and hot-air balloon will both move upwards at a constant velocity.
    When the ball is dropped it continues to move upwards but decelerates constantly (at 9,8 m·s−2) due to the gravitational  attraction force of the earth and slows down until it reaches the highest point in its trajectory (path). It stops momentarily ( v = 0) and then starts to accelerate downwards constantly  (at 9,8 m·s−2). Its speed increases until it hits the ground at a maximum velocity. (8)
  2. The collision between the ball and the ground is inelastic. Some of the ball’s kinetic energy is converted into heat and sound energy and the ball is deformed during the collision. The upward force of the ground on 3 the ball causes it to bounce upwards but the kinetic energy is less than before the collision, so the velocity 3 at which the ball leaves the ground is less than  the velocity at which it hit the ground and the height reached is lower 3 than the previous bounce. (7)
    The point of reference for a position-time graph is placed on the time axis, where y = 0 m.
  3. 5 ms–1 upwards (1)
  4. vi = +5 ms–1 vi = 0 ms–1 a = –9,8 ms–2
    vf2 = vi2 + 2 a ·Δy
    02 = 52 + 2(–9,8)Δy
    –25 = –19,6Δy
    Δ y =  -25   = 1,28 m
             -19.6
    ∴ the ball will reach a maximum height of (60 + 1,28) = 61,28 m above the ground. (5)
  5. vi =+5 ms–1 Δt = 3 sa = –9,8 ms–2
    Δy = vi Δt + ½ at2
    ∴ Δy = (5)(3) + ½ (–9,8)(3)2
    ∴ Δy = – 29,1 m
    ∴ the ball is 29,1 m below the point from where it was released, or (60 – 29,1) = 30,9 m above the ground. (6)
  6. The hot-air balloon moved upwards at a constant velocity.
    Δy = vi Δt + ½ a Δt2
    Δy = (5)(3) + 0
    Δy = 15 m
    ∴ After 3s the hot-air balloon will be 15 m above the starting point.
    We know from Question 4 that the ball will be 29,1 m below the starting point after 3s.
    ∴ after 3s the hot-air balloon and the ball will be (15 + 29,1)
    = 44,1 m apart. (7)
  7. NOTE: Always calculate the velocity at which the ball hits the ground first.
    vi = + 5 ms–1 Δy = – 60 ma = –9,8 ms–2
    vf2 = vi2 + 2a ·Δy 
    vf2 = (5)2 + 2(–9,8)(–60) 
    ∴ vf = 34,66 ms–1 downwards 
    then vf = vi + a ·Δt
    –34,66 = 5 + (–9,8)Δt
    ∴ Δt = 4,05 s (7)
  8. NOTE: The up and down displacement of the ball from the first to the second contact with the ground, is the same in magnitude (size) and tup = tdown.
    Consider the downward motion as negative as in the previous calculations. For the downward part of the bounce:
    17
    vi = 0 ms–1
    Δy = – 8 m
    a = – 9,8 ms–2
    Δy = vi Δt + ½ a Δt2
    – 8 = 0 + ½ (–9,8)(Δt2)
    Δt = 1,28 s
    but tup = tdown
    ∴ the ball takes 1,28 s to reach a height of 8 m. (10)
  9. vf = vi + a Δt
    vf = 0 + (–9,8)(1,28)
    vf = –12,54 ms–1
    ball hits the ground at 12,54 ms–1 downwards after the first bounce (4)

  10. 18(5)

  11. 19(5)

  12. 20(4)
    [69]


Activity 5

  1. A 30 kg iron sphere and a 10 kg aluminium sphere with the same diameter fall freely from the roof of a tall building. Ignore the effects of friction. When the spheres are 5 m above the ground, they have the same ...
    1. momentum.
    2. Acceleration
    3. kinetic energy
    4. potential energy (2)
  2. An object is thrown vertically into the air at 12 m·s–1 in the absence of air friction. When the object is at the highest point, the velocity of the object in m·s–1 is?
    1. 0
    2. 9,8 downwards
    3. 9,8 upwards
    4. 12 (2)
  3. An object is projected vertically upwards and then falls back to the ground level. The acceleration of the object is ….
    1. Directed upwards throughout its movement.
    2. Zero at the greatest height.
    3. Directed downwards throughout its movement.
    4. Directed upwards and then downwards. (2)
      [6]
Solutions
1. B  (2)
2. A  (2)
3. C  (2)
[6]


Activity 6
A ball of mass 0,15 kg is thrown vertically downwards from the top of a building to a concrete floor below. The ball bounces off the floor. The velocity versus time graph below shows the motion of the ball. Ignore the effects of air friction. TAKE DOWNWARD MOTION AS POSITIVE.
21

  1. From the graph, write down the magnitude of the velocity at which the ball bounces off the floor. (1)
  2. Is the collision of the ball with the floor ELASTIC or INELASTIC? refer to the data on the graph to explain the answer. (2)
  3. Calculate the:
    1. Height from which the ball is thrown (3)
    2. Size of the displacement of the ball from the moment it is thrown until time t (5)
      (in an exam, you might see the word “magnitude” — this means “size”).
  4. Sketch a position versus time graph for the motion of the ball from the moment it is thrown until it reaches its maximum height after the bounce. USE THE FLOOR AS THE ZERO POSITION.
    Indicate the following on the graph:
    • The height from which the ball is thrown
    • Time t (4)
      [15]

Solutions

  1. 15 m·s–1 (1)
  2. Inelastic 
    The speed/velocity at which the ball leaves the floor is less / different than that at which it strikes the floor OR The speed/ velocity of the ball changes during the collision. 
    Therefore the kinetic energy changes/is not conserved. (2)
    1. vf2 = vi2 + 2aΔy 
      (20)2 = (10)2 + 2(9,8)Δy 
      ∴ Δy = 15,31 m  (3)
    2. Displacement from floor to maximum height
      vf2 = vi2 + 2aΔy
      (0)2 = (–15)2 + 2(9,8)Δy 
      Δy = –11,48 m 
      Total displacement
      = –11,48 + 15,3 
      = 3,82 m or 3,83 m (5)

  4. 22
    Marking criteria for graph:
    Correct shape as shown for first part.    √ 
    Correct shape as shown for the second part up to t / 2,55 s.  √ 
    Graph starts at –15,3 m at t = 0 s.    √ 
    Maximum height after bounce at time t / 2,55 s.   √ 
    Maximum height after bounce less than 15,3 m.   √ 
    (4)
    [15] 
Last modified on Tuesday, 21 September 2021 12:52
Published in Physical Sciences Paper 1 Grade 12 Study Guide and Notes

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