A-level Mathematics/OCR/M2/Projectile Motion

A projectile is a body which moves under the action of gravity, and may also move horizontally. For M2 we only consider projectiles moving in one plane.

A projectile at launch can be thought as a vector, as either a velocity and angle or initial velocities in x and y planes, shown here.

As the object travels in only one plane, if we assume no air resistance (as is standard in M2), then the projectiles horizontal motion will remain constant while the vertical will only have the objects mass (gravity) acting upon it. This leads to the two equations below.

$$\ x_v = u cos( \theta )\!$$ $$\ y_v = u sin( \theta ) - g t \!$$

Below are the equations for the projectiles displacement. One can derive these from our equations of motion from M1 or integrate with respect to t the velocity equations as $$\ x_v = \frac{d (x_s)}{dt)} \equiv \int x_v dt = x_s + c \!$$ since we define our start point as the origin, c = 0

$$\ x_s = u cos( \theta ) t \!$$ $$\ y_s = u sin( \theta ) t - \tfrac{1}{2} g t^2 \!$$

With these four equations, and applying some facets of symmetry we can solve many novel problem.


 * When $$\ y_s = 0 \!$$, the projectile has just been launched or is in line with the point of launch (or on flat ground, landed). To calculate range, and the time of flight, we use this fact and rearrange

$$\ y_s = u sin( \theta ) t - \tfrac{1}{2} g t^2 \!$$

$$\ 0 = t ( u sin( \theta )  - \tfrac{1}{2} g t ) \!$$

$$\ t = 0\ and\ t = \frac{ 2 u sin( \theta )}{g} \!$$

The x_s at this point in time is the range,

$$\ x_s = u cos( \theta )\ *\ \frac{ 2 u sin( \theta )}{g} \!$$

C3 identity

$$\sin(2x) = 2 \sin (x) \cos(x)\,$$

$$\ x_\triangle = \frac{u^2 sin ( 2 \theta ) }{g} \!$$


 * When $$\ y_v = 0, y_s = maximum \!$$, since our vertical velocity is being accelerated downwards by the object's weight, using M1 knowledge (h being maximum height)

$$\ v^2 = u^2 + 2 a s \!$$

$$\ 0^2 = (u sin( \theta ))^2 + 2 ( - g ) h \!$$

$$\ 0 = u^2 sin^2 ( \theta ) - 2 g h \!$$

$$\ 2 g h = u^2 sin^2 ( \theta ) \!$$

$$\ h_\triangle = \frac{u^2 sin^2 ( \theta )}{2 g} \!$$