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PHYSICS, M5 2025 HSC 28

A bicycle rider jumps from one ramp to a second ramp separated by 16 m as shown. The ramps are inclined at 20° and are 2 m high.
 

What minimum speed is required for the rider to land on the second ramp?   (4 marks)

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\(u=15.6 \ \text{ms}^{-1}\)

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\(\text {Find time of flight}\ \left(S_y=0\right):\)

\(S_y\) \(=u\, \sin \theta\, t+\dfrac{1}{2} a t^2\)
\(0\) \(=u\, \sin 20^{\circ} t-4.9 t^2\)
\(0\) \(=t\left(u\, \sin 20^{\circ}-49 t\right)\)
\(t\) \(=\dfrac{u\, \sin 20^{\circ}}{4.9}\)
♦ Mean mark 47%.

\(\text{Since flight distance = 16 m:}\)

\(S_x\) \(=u \, \cos \theta \ t\)
\(16\) \(=\dfrac{u \,\cos 20^{\circ} \times u \, \sin 20^{\circ}}{4.9}\)
\(u^2\) \(=\dfrac{16 \times 4.9}{\cos 20^{\circ} \times \sin 20^{\circ}}\)
\(u\) \(=\sqrt{\dfrac{16 \times 4.9}{\cos 20^{\circ} \times \sin 20^{\circ}}}=15.6 \ \text{ms}^{-1}\)

PHYSICS, M5 2024 HSC 9 MC

Object \(P\) is dropped from rest, and object \(Q\) is launched horizontally from the same height.
 

Which option correctly compares the projectile motion of \(P\) and \(Q\) ?

  1. The acceleration of \(P\) is less than the acceleration of \(Q\).
  2. The final velocity of \(Q\) is greater than the final velocity of \(P\).
  3. The time of flight of \(Q\) is greater than the time of flight of \(P\).
  4. The initial vertical velocity of \(P\) is less than the initial vertical velocity of \(Q\).
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\(B\)

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  • Both objects start with the same initial vertical velocity and experience the same acceleration due to gravity. Therefore both objects will hit the ground at the same time with the same final vertical velocity.
  • Object \(Q\) however, has both vertical and horizontal velocity while object \(P\) only has vertical velocity. Therefore object \(Q\) will have a greater total final velocity when compared with \(P\).

\(\Rightarrow B\)

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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The position of the ball relative to the launcher’s new position is 44.19 m, 5.2\(^{\circ}\) below the horizontal line of the launcher.

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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 (\(t_1)= 2 \times t_2= 1.167\ \text{s}\)
 

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 \(\frac{1}{2}\) 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.
     

 

\(D\) \(=\sqrt{44.0^2+4^2}=44.18\ \text{m}\)  
\(\theta\) \(=\tan ^{-1}\left(\dfrac{4}{44.0}\right)=5.2^{\circ}\)  

 

  • The final position of the ball relative to \(L\) is 44.18 m, 5.2\(^{\circ}\) below the horizontal line at \(L\).

PHYSICS, M5 2020 HSC 24

The graph shows the vertical displacement of a projectile throughout its trajectory. The range of the projectile is 130 m.
 

Calculate the initial velocity of the projectile.   (4 marks)

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`37\ text{m s}^(-1)`, at 54° above the horizontal.

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From the graph, at  `t=3`, the projectile reaches a maximum height of 44 m:

`s_(y)` `=u_(y)t+(1)/(2)a_(y)t^(2)`  
`44` `=u_(y)(3)-(1)/(2)(9.8)(3^(2))`  
`u_(y)` `=29.4\ text{m s}^(-1)`  

  
Find `u_x` given time of flight = 6 s:

`u_(x)=(s_(x))/(t)=(130)/(6)=21.7\ text{m s}^(-1)`
 

Using Pythagoras:

`u^(2)` `=u_(x)^(2)+u_(y)^(2)`  
  `=21.7^(2)+29.4^(2)`  
`u` `=37\ text{m s}^(-1)`  

  
Find launch angle (`theta)`:

`tan theta` `=(u_y)/(u_x)`  
`theta` `=54^(@)`  

  
So,  `u=37\ text{m s}^(-1)`, at 54° above the horizontal.