A projectile is launched from the ground at an angle of 39° and at a speed of 25 m s\(^{-1}\), as shown in Figure 10. The maximum height that the projectile reaches above the ground is labelled \(h\). --- 5 WORK AREA LINES (style=lined) --- --- 5 WORK AREA LINES (style=lined) ---
a. \(v_v=25\,\sin39^{\circ}=15.733\ \text{ms}^{-1}\) \(\text{At max height:}\ v_v=0\) b. \(\text{Since path is symmetrical:}\) \(\text{Time of flight}\ (t_2) =1.6 \times 2=3.2\ \text{s}\). \(v_h=25\,\cos39^{\circ}=19.43\ \text{ms}^{-1}\ \ \text{(see part (a) diagram)}\) \(\therefore R=v_ht_2=19.43 \times 3.2= 62\ \text{m}\)
\(v\)
\(=u+at_1\)
\(0\)
\(=15.733 -9.8t_1\)
\(9.8t_1\)
\(=15.733\)
\(t_1\)
\(=1.6\ \text{s (2 sig.fig)}\)
PHYSICS, M5 2021 VCE 9-10 MC
Lucy is running horizontally at a speed of 6 m s\(^{-1}\) along a diving platform that is 8.0 m vertically above the water.
Lucy runs off the end of the diving platform and reaches the water below after time \(t\).
She lands feet first at a horizontal distance \(d\) from the end of the diving platform.
Question 9
Which one of the following expressions correctly gives the distance \(d\) ?
- 0.8\(t\)
- 6\(t\)
- 5\(t^2\)
- 6\(t\) + 5\(t^2\)
Question 10
Which one of the following is closest to the time taken, \(t\), for Lucy to reach the water below?
- 0.8 s
- 1.1 s
- 1.3 s
- 1.6 s
PHYSICS, M5 2023 HSC 32
A horizontal disc rotates at 3 revolutions per second around its centre, with the top of the disc at ground level.
At 2 m from the centre of the disc, a ball is held in place at ground level on the top of the disc by a spring-loaded projectile launcher. At position \(X\), the launcher fires the ball vertically upward with a velocity of 5.72 m s\(^{-1}\).
Calculate the ball's position relative to the launcher's new position, at the instant the ball hits the ground. (7 marks)
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Show Worked Solution
Find horizontal velocity of the ball, \(v_{\text{x}}\):
\(T=\dfrac{1}{3} = 0.333\ \text{seconds} \)
| \(v_{\text{x}}\) | \(=\dfrac{2\pi r}{T}\) | |
| \(=\dfrac{2\pi \times 2}{0.333…}\) | ||
| \(=37.699\ \text{ms}^{-1}\) |
♦ Mean mark 53%.
Calculating the time of flight, \(t_1\):
→ Let \(t_2\) = time to max height
| \(v_{\text{y}}\) | \(=u_{\text{y}} + at_2\) | |
| \(t_2\) | \(=\dfrac{v_{\text{y}}-u_{\text{y}}}{a}\) | |
| \(=\dfrac{0-5.72}{-9.8}\) | ||
| \(=0.58367\ \text{sec} \) |
→ Time of flight (\(t_1\)) = 2 × (\(t_2\)) = 1.167 s
Range of the ball from launch position:
| \(s_{\text{x}}\) | \(=v_{\text{x}} \times t_2\) | |
| \(=37.699 \times 1.167\) | ||
| \(=44.0\ \text{m}\) |
Position of the launcher (L) when the ball hits the ground:
→ Revolutions (before ball lands) = 3 × 1.167 = 3.5 revolutions
→ The Launcher (L) is \(\dfrac{1}{2}\) a revolution past its starting point.
→ Thus, the positions of both the ball and the launcher at the time when the ball hits the ground can be demonstrated in the diagram below.
| \(D\) | \(=\sqrt{44.0^2+4^2}\) | |
| \(=44.18\ \text{m}\) | ||
| \(\theta\) | \(=\tan ^{-1}(\dfrac{4}{44.0})\) | |
| \(=5.2^{\circ}\) |
→ The final position of the ball relative to \(L\) is 44.18 m, 5.2\(^{\circ}\) below the horizontal line at \(L\).
PHYSICS, M5 2023 HSC 8 MC
A ball is launched from a platform at position \(A\) with velocity \(u\). It lands in the position shown.
The ball could be made to land at position \(B\) by increasing the
- velocity \(u\).
- launch angle.
- mass of the ball.
- height of the platform.
Show Worked Solution
→ As the launch angle increases, the horizontal velocity of the ball decreases.
→ At a steep launch angle, the ball will travel high into the air but have a short range, thus it could be made to land at position \(B\) as seen in the diagram below.
\(\Rightarrow B\)
Mean mark 56%.
PHYSICS, M5 EQ-Bank 26
A baseball is hit with a velocity of 28 m s ¯1 at an angle of 30° to the horizontal at an initial height of 1.0 m above the plate. Ignore air resistance in your calculations.
- How long does it take the ball to return to the initial height above the ground? (3 marks)
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- The ball is hit directly towards a stationary outfielder who is 85 m from the plate. At the instant the ball is hit, the outfielder begins to run towards the plate with constant acceleration.
- What is the magnitude of her acceleration if she catches the ball when it is 0.50 m above the ground? (4 marks)
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PHYSICS, M5 EQ-Bank 1 MC
Some students were testing the hypothesis that launching a projectile at an angle of 45° will give the maximum horizontal range.
Which experimental setup will best test the hypothesis?
PHYSICS, M5 2017 HSC 29
A spring is used to construct a device to launch a projectile. The force `(F)` required to compress the spring is measured as a function of the displacement `(x)` by which the spring is compressed.
The potential energy stored in the compressed spring can be calculated from `E_p=(1)/(2) kx^(2)`, where `k` is the gradient of the force-displacement graph shown.
- A projectile of mass 0.04 kg is launched using this device with the spring compressed by 0.08 m. Calculate the launch velocity. (4 marks)
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- Calculate the range of a projectile launched by this device from ground level at an angle of 60° to the horizontal with a velocity of 10 `text{m s}^(-1)`. (3 marks)
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PHYSICS, M6 2020 HSC 34
A charged particle, `q_1`, is fired midway between oppositely charged plates `X` and `Y`, as shown in Figure 1. The voltage between the plates is `V` volts.
The particle strikes plate `Y` at point `P`, a horizontal distance `s` from the edge of the plate. Ignore the effect of gravity.
Plate `Y` is then moved to the position shown in Figure 2, with the voltage between the plates remaining the same.
An identical particle, `q_2`, is fired into the electric field at the same velocity, entering the field at the same distance from plate `X` as `q_1`.
- Compare the work done on `q_1` and `q_2`. (3 marks)
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- Compare the horizontal distances travelled by `q_1` and `q_2` in the electric field. (3 marks)
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