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As we have seen in previous lessons, projectiles can be broken into vertical and horizontal components.

- The horizontal component of initial velocity
- U
_{x }= u cos θ

- U
- The vertical component of initial velocity
- U
_{y }= u sin θ

- U

Projectile motion can be modelled using the following equations:

To use these as x and y components, remember that the acceleration in the x plane is 0 and the acceleration in the y plane is g (9.8 m/s/s)

We derive the following equations for the y direction

- v
_{y }= u_{y }+ gt - s
_{y }= u_{y}t + ½ g t^{2} - v
_{y}^{2}= u_{y}^{2 }+ 2 g s

As well as the following for the x direction

- v
_{x }= u_{x} - s
_{x }= u_{x}t - v
_{x}^{2}= u_{x}^{2}

Memorize.

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Master.

The following will be HSC style projectile motion questions.

Question 1 a. [1 mark]

A potato launcher releases potato’s at an angle of 35° to the horizontal, it’s muzzle velocity is 400 m/s. What are the initial horizontal and vertical components of the velocity of the potato?

Question 1 b. [2 marks]

What are the vertical and horizontal components of the potato 5 seconds after it has been launched?

Question 1 c. [3 marks]

What is the range of the potato launcher?

Question 2. [3 marks]

A tennis ball is hit 1.5 m from the ground, it reaches a maximum height of 36.5 m with an initial angle of 60°, what was it’s initial speed?

Question 3 a. [3 marks]

A tennis ball is thrown from the top of the wall where with a launch height of 12 m above the ground. It has an initial velocity of 12 m/s at an angle of 40° from the horizontal. How long is the ball in the air for?

Question 3 b. [3 marks]

What would the final velocity of the tennis ball be before it hits the ground?

Question 4. [3 mark]

An archer is 20 m away from a target, he shoots his arrow at 60 km/ h. What angle must he aim at to hit the target?

Question 5. [3 marks] (extension)

A ball is thrown off a cliff 30 degrees up from the horizontal. If the ball lands 60 m from the cliff, where the height of the cliff is 4.2 m, with a final velocity of 15 m/s 110 degrees from the horizontal. What is the initial velocity of the ball?

Answers.

Question 1.

a)

Ux = 330 m/s (2 sig figs)

uy = 230 m/s (2 sig figs)

b)

Vx = Ux = 330 m/s (2 sig fig)

Vy = uy + gt = 229 – 9.8*5 = 180 m/s

c)

15000 m

Question 2.

30 m/s

Question 3.

a) 2.5 sec

b) 13 at 175 degrees

Question 4.

- Approach this by measuring how far the arrow falls if it was shot horizontally
- Using trig determine the angle it fell
- Therefore, shooting the arrow up at this angle will result in it falling to the correct height – this is because the angle is so small so negligible amount of energy is lost in the y direction.

20 degrees up from horizontal

Question 5.

v_{x} = 15 * sin (70)

v_{x} = u_{x}

u_{x} = u cos 30 = 15 * cos (70)

u = 15 * sin (70)/ cos (30) = 75 m/s