# PHY C3: Projectile Motion

Now we’ll be covering:

• Two-dimensional Motion under a Constant Force (Projectile Motion)
• Characteristics
• Problem-solving

Let’s jump in!

What is Projectile Motion?
Two-dimensional Motion under a Constant Force.
So far we’ve been learning about motion in ONE dimension (linear motion).

What if there are 2 dimensions of motion?
In this case, there are 2 components that have to be taken into account.

In projectile motion, there is a CONSTANT FORCE being applied in ONE of the components.

A few examples:

 An object thrown at a certain angle into the air, which then falls The object is acted upon by a constant GRAVITATIONAL FORCE in the VERTICAL component An object dropped from a horizontally-moving platform A particle moving at an angle in an electric field The particle is acted upon by a constant ELECTRIC FORCE from one plate to another

For the next part, we’ll use the first example (an object thrown at an angle).

Characteristics of Projectile Motion

What’s happening here?

1. The object is thrown with a FORCE at an ANGLE
2. This force accelerates the object from rest to a certain velocity, u
(initial velocity in resultant direction of motion)
3. When the object leaves the hand, it NO LONGER has a force applied in the horizontal direction
(& thus NO horizontal acceleration)
4. Throughout flight, the object is ONLY affected by GRAVITATIONAL FORCE
(it is in FREE FALL)
5. The object travels in a parabola until it hits the ground

 Aspect Characteristics Horizontal Motion NO external force applied Constant Velocity 0 Acceleration Vertical Motion Affected by Gravitational Force Increasing Velocity (downwards) Constant Acceleration (g = 9.81 ms-2) Resultant Motion NOT constant velocity The resultant of the x & y components at any given point (tangent to curve)

Now that you have these in mind, let’s see what you can calculate in this situation:

 Horizontal Motion (x-axis) Velocity stays constant throughout flight ux = vxNo acceleration Vertical Motion (y-axis) Velocity starts as positive, decreases, becomes 0 at max height, becomes negative. Acceleration is constant: -g = 9.81 m/s2 downwards It’s helpful to split the parabola into 2 parts: 1ST HALF: object going up 2ND HALF: object going down Going up, the initial v: uy1 = u sin θ the final v: vy1 = 0 Because it has reached max height. Going down, the initial v: uy2 = 0 the final v: vy2 = u sin θ (Remember to add a + or – sign to denote whether it’s travelling up or down!) Velocity in Direction of Motion Resultant of V & H components of velocity at any given pointv = vx / cos θ v = vy / sin θ Max Height At max height, vy = 0 Max Height can be found by considering the 1ST HALF of the flight (from ground to max height): Initial Vertical Velocity (UPWARDS) uy1 = v sin θ t1 = time of flight / 2 & understanding that: Final V of 1ST HALF vy1 = Initial V of 2ND HALF uy2 = 0 a = g = – 9.81 ms-2 Formulas to use, based on info given: s = ½ uy1t1 s = uy1t1 + ½ gt12 0 = uy12 + 2gs These are just the SUVAT formulas, just sub in the values you know. Since the Parabola is symmetrical, the exact same methods can be used for the 2ND HALF of the flight (from Max Height to ground). Just flip the signs for acceleration & velocity (since the object is travelling DOWNWARDS). Range Horizontal distance travelled by object. If you know the TIME OF FLIGHT, you can find range by: s = uxt (since ux is constant throughout the flight) Time of Flight Time taken from launch to landing. Since the parabola is SYMMETRICAL, time taken from launch to max height (t1) = time taken from max height to landing (t2) Thus, if you can calculate either t1 or t2, you can find the total time of fight by multiplying by 2.

Here’s a little game you can play to explore more about projectile motion:
PhET Simulation: Projectile Motion

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