Projectile motion

[Bri MT]

In units 1&2 you should’ve gained some practice in 1D motion – now in units 3&4 we’re going to look at 2D motion. The way we’ll approach this is by splitting the 2D motion into 1D components, and using the same constant acceleration formulas we’re used to.


Components of velocity
Typically, a question on this will involve a projectile launched at some initial velocity u with angle θ.
First, we’ll want to split u into horizontal (I'm denoting this with x) and vertical (I'm denoting this with y) components.
u_y = usin(θ)
u_x = ucos(θ)

Other info
Now that we have the u_x and u_y it’s time to think about what other information we have. In QCE physics you assume that there is no drag/air resistance, so for the purposes of our calculations there is no acceleration in the horizontal direction. We also know that gravity exists (I hope you do anyway) and we’re going to assume it’s constant since presumably this is occurring near Earth’s surface.

This is summed up below:  (I recommend writing this down when you are answering a question)
The horizontal / x direction:
u = ucos(θ)
v = u = ucos(θ)
a = 0

The vertical/ y direction:
v = 0
u = ucos(θ)
a = -9.8 ms^-2
Be careful here! You can define up as positive like I have done or as negative and you must make sure the initial velocity and acceleration are consistent with your choice.

Finding time
The y direction dictates our time of flight, and given the values we have we can sub and solve for t when the velocity is zero:
v_y = u_y + gt
=> u_y/9.8 = t   note that because we chose v=0 this t is the time taken to reach the peak of the flight

If our flight is symmetrical (i.e. launching and landing at the same height) we can multiply this by 2 to get the full time of flight. Time of flight is the same for both the horizontal and vertical components. Otherwise, you’ll need to use the constant acceleration formulas to find this (remember: use the vertical direction to find time)

Finding displacements
Finding the range (horizontal displacement between starting and landing point) is very simple: just multiply the full time of flight by the horizontal component of the velocity.

To find the maximum vertical displacemente we use the time taken to reach zero velocity (reminder: in a symmetrical flight this is half of the time of flight)
sy = .5gt² + u_yt

If you have any questions, if you have a method that works well for you, or anything like that please feel free to share To do so you'll need to register for free, which you can do here