PROJECTILE MOTION SIMULATION

The Calculations

We are are repeatdly applying the constant acceleration equations to get updating values for the horizontal and vertical displacement and the vertical velocity. Note, horizontal velocity is constant since there is no horizontal acceleration.

We will use the following SUVAT equations to update out position and velocity: $$\small{ s_x = v_x \times t }$$ $$ \small{ s_y =u_y \times t - \frac{1}{2} \times -9.81 \times t^2 }$$ $$ \small{ v_y = u_y - 9.81 \times t }$$

The inital conditions for this projectile are given by (chosen by you!):

Initial height: , Initial Vertical Velocity: = u$$_y, Initial Horizontal Velocity: $$ = v_x

So we can calculate the new x position s$$_x=v_xt = * = m

Also we can calculate the new y position s$$_y = u_y*t - 0.5*9.81*t² + initial height = * - 0.5 * 9.81 * + = m

Finally, we can calculate the new y velocity v$$_y = u_y - 9.81*t = - 9.81 * = m/s

Where s$$_x is the horizontal position, s$$_y is the vertical position, v$$_x is the horizontal velocity, v$$_y is the vertical velocity, u$$_y is the initial vertical velocity, and t is the time since launch.

We can also calculate the vertical & horizontal range and the time of flight:

Now check your understanding by calculating the vertical and horizontal range! (round your answers to the nearest whole number)

The vertical range is given by $ \small verticalRange = {\frac{u_y^2}{2 \times 9.81} + initialHeight } $

Using the known values and the formula provided calculate the vertical range: metres .

Correct!

Incorrect, the answer is vertical range ≈ $$² / (2*9.81) + ≈

The time of flight is given by $ \small{t =\frac{-u_y-\sqrt{-u_y-2\times (-9.81)\times initialHeight}}{-9.81}} $

Using the known values and the formula provided calculate the time of flight: seconds .

Correct!

Incorrect, the answer is time of flight ≈ (- $$ - √  -²- 2*(-9.81)* )/(-9.81) ≈

The horizontal range is given by $ \small{ horizontalRange = u_x \times timeofflight } $

Using the known values and the formula provided calculate the horizontal range: metres .

Correct!

Incorrect, the answer is Horizontal Range ≈ * ≈

Note, as with all projectile motion examples we are making a lot of assumptions: (1) there is only vertical acceleration which is gravity (2) we are taking the earth as flat (3) each axis is a 'wall' (4) the particle does not bounce