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A man throws ball A downwards with a speed of 2 m s⁻¹ from the edge of a window, 45 m above a dam of water - NSC Physical Sciences - Question 3 - 2016 - Paper 1

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A man throws ball A downwards with a speed of 2 m s⁻¹ from the edge of a window, 45 m above a dam of water. One second later he throws a second ball, ball B, downwar... show full transcript

Worked Solution & Example Answer:A man throws ball A downwards with a speed of 2 m s⁻¹ from the edge of a window, 45 m above a dam of water - NSC Physical Sciences - Question 3 - 2016 - Paper 1

Step 1

3.1.1 Speed with which ball A hits the surface of the water

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Answer

To find the speed at which ball A hits the water, we can use the kinematic equation:

v_f^2 = v_i^2 + 2aigg(\Delta y\bigg)

where:

  • vfv_f is the final velocity,
  • vi=2v_i = 2 m/s (initial velocity of ball A),
  • a=9.8a = 9.8 m/s² (acceleration due to gravity),
  • Δy=45\Delta y = -45 m (displacement, negative since it's downwards).

Plugging the numbers in: vf2=(2)2+2(9.8)(45)v_f^2 = (2)^2 + 2(9.8)(-45) vf2=4882v_f^2 = 4 - 882 vf2=878v_f^2 = -878

Taking the square root gives:
vf=extsqrt(878)ext(Ignoresignsincespeedispositive)v_f = ext{sqrt}(878) ext{ (Ignore sign since speed is positive)} Thus, vf29.6extm/sv_f ≈ 29.6 ext{ m/s}.

Step 2

3.1.2 Time it takes for ball B to hit the surface of the water

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Since ball B is thrown one second after ball A, we denote the time for ball A as tAt_A. The time for ball B would then be tB=tA1t_B = t_A - 1 s. To calculate the time for ball A, we use:

Δy=vit+12at2\Delta y = v_i t + \frac{1}{2} a t^2

For ball A: 45=2tA+12(9.8)tA2-45 = 2t_A + \frac{1}{2} (9.8) t_A^2

This is a quadratic equation in the form: 4.9tA2+2tA+45=04.9t_A^2 + 2t_A + 45 = 0

Using the quadratic formula: tA=b±b24ac2at_A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} With a=4.9a = 4.9, b=2b = 2, and c=45c = 45:

Calculating: tA=2±(2)24(4.9)(45)2(4.9)t_A = \frac{-2 \pm \sqrt{(2)^2 - 4(4.9)(45)}}{2(4.9)}

This results in a positive value approximately: tA2.83st_A ≈ 2.83s Thus, the time for ball B to hit the surface is: tB=2.831=1.83st_B = 2.83 - 1 = 1.83 s.

Step 3

3.1.3 Initial velocity of ball B

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To find the initial velocity of ball B, we note that it drops one second after ball A. Since both balls hit the water at the same time and the only variable change is the time of release, we can say:

Since ball A’s total time is 2.83s2.83 s while ball B’s is 1.83s1.83 s, we can use the same kinematic equation:

vf=vi+atv_f = v_i + at

For ball B, we have: 29.6=vi+(9.8)(1.83)29.6 = v_i + (9.8)(1.83)

Solving for viv_i gives: vi=29.6(9.8)(1.83)v_i = 29.6 - (9.8)(1.83)
vi15.62m/sv_i ≈ 15.62 m/s.

Step 4

3.2 On the same set of axes, sketch a velocity versus time graph for the motion of balls A and B

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The graph will have time on the x-axis and velocity on the y-axis. For ball A, the initial velocity is 2 m/s and it accelerates downwards due to gravity, leading to a straight line with a positive slope.

For ball B, its initial velocity is determined to be about 15.62 m/s and it also accelerates downwards but starts one second later. The graph for ball B will start at time 1s on the x-axis with a higher initial velocity compared to ball A.

Features to indicate on the graph:

  • Mark the initial velocities at the respective time points.
  • Indicate that both lines intersect at the point where they hit the water,
  • Clearly label time at which ball B is released (t = 1s) and when both balls reach the water.

For clearer illustration, ensure the y-values indicate their respective velocities at their points of intersection.

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