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PHYSICS, M5 2019 VCE 10

A projectile is launched from the ground at an angle of 39° and at a speed of 25 m s\(^{-1}\), as shown in the diagram. The maximum height that the projectile reaches above the ground is labelled \(h\).
 

  1. Ignoring air resistance, show that the projectile's time of flight from the launch to the highest point is equal to 1.6 seconds. Give your answer to two significant figures. Show your working and indicate your reasoning.  (2 marks)

  2. Calculate the range, \(R\), of the projectile. Show your working.  (2 marks)

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a.    \(t=1.6\ \text{s}\)

b.    \(62\ \text{m}\)

Show Worked Solution

a.    
       

\(v_v=25\,\sin39^{\circ}=15.733\ \text{ms}^{-1}\)

\(\text{At max height:}\ v_v=0\)

\(v\) \(=u+at_1\)  
\(0\) \(=15.733 -9.8t_1\)  
\(9.8t_1\) \(=15.733\)  
\(t_1\) \(=1.6\ \text{s  (2 sig.fig)}\)  

 

b.    \(\text{Since path is symmetrical:}\)

\(\text{Time of flight}\ (t_2) =1.6 \times 2=3.2\ \text{s}\).

\(v_h=25\,\cos39^{\circ}=19.43\ \text{ms}^{-1}\ \ \text{(see part (a) diagram)}\) 

\(\therefore R=v_ht_2=19.43 \times 3.2= 62\ \text{m}\)

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 \(d\) ?

  1. 0.8\(t\)
  2. 6\(t\)
  3. 5\(t^2\)
  4. 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?

  1. 0.8 s
  2. 1.1 s
  3. 1.3 s
  4. 1.6 s
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\(\text{Question 9:} \ B\)

\(\text{Question 10:} \ C\)

Show Worked Solution

Question 9

  • The horizontal displacement formula for projectile motion:  \(s=ut\)
  • Horizontal distance: \(d=6t\)

\(\Rightarrow B\)
 

Question 10

Find time of flight \((t):\)

\(s\) \(=ut+\dfrac{1}{2}at^2\)  
\(s\) \(=\dfrac{1}{2}at^2\ \ \ (u=0)\)  
\(t\) \(=\sqrt{\dfrac{2s}{a}}=\sqrt{\dfrac{2 \times 8}{9.8}}\approx 1.3\ \text{s}\)  

 

\(\Rightarrow C\)

PHYSICS, M5 2023 VCE 9

Giorgos is practising his tennis serve using a tennis ball of mass 56 g.

  1. Giorgos practises throwing the ball vertically upwards from point A to point B, as shown in Diagram A. His daughter Eka, a physics student, models the throw, assuming that the ball is at the level of Giorgos's shoulder, point A, both when it leaves his hand and also when he catches it again. Point A is 1.8 m from the ground. The ball reaches a maximum height, point B, 1.8 m above Giorgos's shoulder.

  

  1. Show that the ball is in the air for 1.2 s from the time it leaves Giorgos's hand, which is level with his shoulder, until he catches it again at the same height.   (2 marks)
  1. Giorgos swings his racquet from point D through point C, which is horizontally behind him at shoulder height, as shown in Diagram B, to point B. Eka models this swing as circular motion of the racquet head. The centre of the racquet head moves with constant speed in a circular arc of radius 1.8 m from point C to point B.

  1. The racquet passes point C at the same time that the ball is released at point A and then the racquet hits the ball at point B.
  2. Calculate the speed of the racquet at point C.   (2 marks)
  1. The ball leaves Giorgos's racquet with an initial speed of 24 m s\(^{-1}\) in a horizontal direction, as shown in Diagram C. A tennis net is located 12 m in front of Giorgos and has a height of 0.90 m.
     

  1. How far above the net will the ball be when it passes above the net? Assume that there is no air resistance. Show your working.   (3 marks)

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a.    \(1.2\ \text{s}\)

b.    \(4.7\ \text{ms}^{-1}\)

c.    \(1.475\ \text{m}\)

Show Worked Solution

a.    Time to for the ball to fall from max height back to shoulder level:

\(s\) \(=ut+\dfrac{1}{2}at^2\)  
\(1.8\) \(=0 \times t + \dfrac{1}{2} \times 9.8 \times t^2\)  
\(t\) \(=\sqrt{\dfrac{1.8}{4.9}}=0.606\ \text{s}\)  

 

\(\text{Total time of flight}\ =0.606 \times 2=1.2\ \text{s}\)
 

b.    The racket travels a quarter of the circumference of the circle in 0.606 seconds.

\(v=\dfrac{d}{t}=\dfrac{2\pi \times 1.8 \times 0.25}{0.606}=4.7\ \text{ms}^{-1}\)
 

♦♦ Mean mark (b) 27%.

c.    Time for the ball to reach the net:

  \(t=\dfrac{d}{v}=\dfrac{12}{24}=0.5\ \text{s}\)

Vertical displacement from max height in 0.5 seconds:

  \(s=ut+\dfrac{1}{2}at^2=0 \times 0.5 + \dfrac{1}{2} \times 9.8 \times 0.5^2=1.225\ \text{m}\).

\(\text{Height (at net)}\ =3.6-1.225=2.375\ \text{m}\).

\(\therefore\ \text{Height (above net)}\ =2.375-0.9=1.475\ \text{m}\)

♦ Mean mark (c) 47%.

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.

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

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 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.
 

  1. How long does it take the ball to return to the initial height above the ground?   (3 marks)
  1. 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.
  2. 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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a.   `t=2.9\ text{seconds}`

b.   `a=3.6  text{m s}^(-2)`

Show Worked Solution

a.   At initial height: `s_y=0`

Initial vertical velocity:  `u_y=28sin30°=14\ text{m s}^(-1)`

Vertical acceleration: `-9.8\ text{m s}^(-2)`

`s_(y)` `=u_(y)t+(1)/(2)a_(y)t^2`  
`0` `=14 t+(1)/(2)xx(-9.8)xxt^(2)`  
`0` `=14 t-4.9 xxt^(2)`  
`0` `=t(14-4.9 t)`  
`t` `=(14)/(4.9)=2.9\ text{seconds},\ \ (t>0)`  

 

b.   At  `s_y=0.50\ text{m}`:

`v_(y)^2` `=u_(y)^2+2a_(y)s_(y)`  
`v_(y)^2` `=14^(2)+2(-9.8)(-0.50)`  
`v_(y)` `=sqrt(205.8)~~14.3457\ text{m s}^(-1)`  

 
Using  `v_y=u_(y)+a_(y)t:`

`-14.3457` `=14-9.8t`  
`t` `=(14+14.3457)/(9.8)=2.892\ text{s}`  

 
Horizontal range travelled (`s_x`):

`s_(x)=u_(x)t=28 cos 30^{\circ} xx 2.892=70.127\ text{m}`
 

COMMENT: Note `v_y` and `u_y` are in opposite directions and have opposite signs.

  Distance fielder travelled `=85-70.127 = 14.873\ text{m}`

`s` `=ut+(1)/(2)at^(2)`  
`14.873` `=0+(1)/(2)xx a xx2.892^(2)`  
`:.a` `=(2 xx 14.873)/2.892^(2)`  
  `=3.6\ text{m s}^(-2)\ \ text{(to 1 d.p.)}`  

PHYSICS, M5 2021 HSC 34

A 3.0 kg mass is launched from the edge of a cliff.
 

The kinetic energy of the mass is graphed from the moment it is launched until it hits the ground at `X`. The kinetic energy of the mass is provided for times `t_0, t_1` and `t_2`.
 

  1. Account for the relative values of kinetic energy at `t_0, t_1` and `t_2`.    (4 marks)
  1. The horizontal component of the velocity of the mass during its flight is 13.76 `text{m s}^(-1)`.    (3 marks)
  2. Calculate the time of flight of the mass.
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a.   Relative values of kinetic energy at `t_0, t_1` and `t_2`:

  • After the mass is launched at ` t_0`, downwards gravitational acceleration decreases the vertical velocity of the mass until it is zero at `t_1`.
  • The kinetic energy of the mass is at a minimum here but not zero due to the horizontal velocity of the mass, which remains constant until the projectile hits the ground.
  • After `t_1`, the kinetic energy of the mass increases as its gravitational potential (GPE) energy is converted to kinetic energy. This occurs as gravity increases the vertical velocity of the mass until it strikes the ground at `t_2`.
  • The GPE of the mass at `t_2` is lower than its GPE at`t_0`.
  • The mass has greater kinetic energy at `t_2` than at `t_0.`

b.   4.8 seconds

Show Worked Solution

a.   Relative values of kinetic energy at `t_0, t_1` and `t_2`:

  • After the mass is launched at ` t_0`, downwards gravitational acceleration decreases the vertical velocity of the mass until it is zero at `t_1`.
  • The kinetic energy of the mass is at a minimum here but not zero due to the horizontal velocity of the mass, which remains constant until the projectile hits the ground.
  • After `t_1`, the kinetic energy of the mass increases as its gravitational potential (GPE) energy is converted to kinetic energy. This occurs as gravity increases the vertical velocity of the mass until it strikes the ground at `t_2`.
  • The GPE of the mass at `t_2` is lower than its GPE at`t_0`.
  • The mass has greater kinetic energy at `t_2` than at `t_0.` 

b.   Find  `u_y` when `t=0`:

`KE` `=(1)/(2)mv^(2)`  
`v^2` `=2 xx 864/(3)=576`  
`v` `=24`  

 
Using Pythagoras:

`24^(2)=u_(y)^(2)+13.76^(2)\ \ =>\ \ u_(y)=19.66\ text{m s}^(-1)`
 
Find  `v_y`  at  `t_2`:`

`1393` `=(1)/(2) xx 3v^(2)`  
`v^2` `=2 xx 1393/(3)=928.7`  
`v` `=30.47`  

 
Using Pythagoras:

`30.47^(2)=v_(y)^(2)+13.76^(2)\ \ =>\ \ v_(y)=27.19\ text{m s}^(-1)`

`v_(y)` `=u_(y)+a_(g)t`  
  `t` `=(27.19-(-19.66))/(9.8)=4.8\ \text{sec}`  

PHYSICS, M5 2020 HSC 5 MC

A student throws a ball that follows a parabolic trajectory.

What change to the initial velocity would make the ball's time of flight shorter?

  1. Increasing only the vertical component
  2. Decreasing only the vertical component
  3. Increasing only the horizontal component
  4. Decreasing only the horizontal component
Show Answers Only

`B`

Show Worked Solution
  • Only the vertical component of the ball’s velocity impacts its time of flight.
  • Decreasing this decreases the time of flight.

`=> B`