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 = uxt
- vx2 = ux2
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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?
Ux = 330 m/s (2 sig figs)
uy = 230 m/s (2 sig figs)
Vx = Ux = 330 m/s (2 sig fig)
Vy = uy + gt = 229 – 9.8*5 = 180 m/s
a) 2.5 sec
b) 13 at 175 degrees
- 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
vx = 15 * sin (70)
vx = ux
ux = u cos 30 = 15 * cos (70)
u = 15 * sin (70)/ cos (30) = 75 m/s
2 thoughts on “Projectile motion calculations”
Would you by any chance have the answer to the extension question? Many thanks!!