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A small disc, C, is thrown vertically upwards at a speed of 15 m·s⁻¹ from the edge of the roof of a building of height 30 m - NSC Physical Sciences - Question 3 - 2022 - Paper 1

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A small disc, C, is thrown vertically upwards at a speed of 15 m·s⁻¹ from the edge of the roof of a building of height 30 m. AFTER 0.5 s, a small ball B is shot vert... show full transcript

Worked Solution & Example Answer:A small disc, C, is thrown vertically upwards at a speed of 15 m·s⁻¹ from the edge of the roof of a building of height 30 m - NSC Physical Sciences - Question 3 - 2022 - Paper 1

Step 1

3.1 Explain the term projectile.

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Answer

A projectile is an object that is thrown into the air with an initial velocity and is influenced only by the force of gravity and its initial thrust. In this context, both disc C and ball B are considered projectiles as they are influenced by the Earth's gravitational pull once thrown.

Step 2

3.2.1 Time taken by disc C to reach its maximum height.

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To find the time taken for disc C to reach its maximum height, we can use the equation of motion:

v=u+atv = u + at

Where:

  • Final velocity, v=0v = 0 m/s (at maximum height)
  • Initial velocity, u=15u = 15 m/s
  • Acceleration due to gravity, a=9.81a = -9.81 m/s² (acting downward)

Setting the equation to zero:

0=159.81t0 = 15 - 9.81t

Rearranging gives:

9.81t=159.81t = 15

Thus, the time:

t=159.811.53extsecondst = \frac{15}{9.81} \approx 1.53 ext{ seconds}

Step 3

3.2.2 Maximum height above the ground reached by disc C.

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To calculate the maximum height, we can use the formula:

h=ut+12at2h = ut + \frac{1}{2}at^2

Substituting for:

  • Initial velocity, u=15u = 15 m/s
  • Time, t=1.53t = 1.53 seconds
  • Acceleration, a=9.81a = -9.81 m/s²:

h=15(1.53)+12(9.81)(1.532)h = 15(1.53) + \frac{1}{2}(-9.81)(1.53^2)

Calculating gives:

h22.95extmetersabovetheroof.h \approx 22.95 ext{ meters above the roof.}

Since the building height is 30 m, the maximum height above the ground is:

30+22.9552.95extmeters30 + 22.95 \approx 52.95 ext{ meters}

Step 4

3.3 Calculate the time from the moment that disc C was thrown upwards until the time ball B hits the disc.

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First, we calculate the total time the disc C took to reach its maximum height (1.53 seconds) and falls back down to the original height (30 m). The total time for the entire journey up and down:

  • Time up = 1.53 seconds

To find the time of descent from 52.95 m back down to 30 m, we can use:

h=ut+12at2h = ut + \frac{1}{2}at^2

Where:

  • Initial height = 52.95 m
  • Final height = 30 m

From 52.95 m to 30 m, descending:

52.9530=9.81t+12(9.81)t252.95 - 30 = 9.81t + \frac{1}{2}(-9.81)t^2

This leads to the quadratic equation:

0=4.905t29.81t+22.950 = 4.905t^2 - 9.81t + 22.95

Solving this using the quadratic formula gives:

  • t2.52t \approx 2.52 seconds.

Total time = 1.53 + 0.5 + 2.52 = 4.55 seconds from when C is thrown.

Step 5

3.4 On the same set of axes, sketch graphs of velocity versus time for disc C and ball B.

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Answer

To sketch the graph:

  1. Initial Velocities:
    • Disc C starts from 15 m/s going upwards.
    • Ball B starts from 0 m/s but is shot upward at 0.5 seconds with 40 m/s.
  2. Time to Max Height for Disc C: The line should show a decrease in velocity until it reaches 0 at 1.53 seconds.
  3. Ball B's Motion: It will have an initial increase until it reaches its maximum point, while its graph should reflect the time it takes to reach disc C, factoring in its higher initial speed.
  4. Points to Indicate: Mark the time at which ball B was shot (0.5 seconds), the time at which disc C reaches its maximum height (1.53 seconds), and when both collide.
  5. Label the two lines distinctly as B (for ball B) and C (for disc C).

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