# Projectile motion calculations

Learn.

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

• The horizontal component of initial velocity
• Ux = u cos θ
• The vertical component of initial velocity
• Uy = u sin θ

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

• vy = uy + gt
• sy = uyt + ½ g t2
• vy2 = uy2 + 2 g s

As well as the following for the x direction

• vx = ux
• sx = ux
• vx2 = ux2

Memorize.

You only need to memorize.

You can derive the other equations from what is on your formula sheet.

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?

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.

vx = 15 * sin (70)

vx = ux

ux = u cos 30 = 15 * cos (70)

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