Using Differential Equations To Solve Missile Trajectory Problems

  A series of separate differential equations can be set up using the diagram shown below, based on projectile motion - incorporating gravity but neglecting air drag effects.

Diagram depicting projectile motion in 2 dimensions.

Here we can write the DEs for time t=0, at position x=0, y = 0:

v _x =  dx/dt =  v_o cos(q)

v_y =  dy/dt  =  v_o sin(q)

The force components at time t are:

F _x =   0   and:   F _y =   - mg

With these incorporated the differential equations become:

i) m d²x/dt ²  =   0   and ii) m d²y/ dt ²  =  - mg

Each equation above requires two integrations, yielding 4 constant of integration in all. 

For eqn. (i) we get on integrating:  dx/ dt = c1   Then:

ò dx  =   c1 òdt

Yields:  x = c1t + c2

Here for eqn. (ii) we have:

d²y/ dt ²  =  - g     So:   dy/dt = - gt  + c3

And:  y =  - ½ g t ²   +  c3t  +  c4

Based on the initial conditions given we have:

c1 =  v_o cos(q),    c2 = 0,   c3 =   v_o sin(q),  c4 =  0

Then the position of the projectile  at time t seconds after firing can be found from:  

x  =  v_o cos(q) t    and:   y = - ½ g t ²   + v_o sin(q) t

Clearly, the projectile attains maximum altitude when its y -component of velocity is 0, i.e.

dy/ dt = 0 =  -gt  +  v_o sin(q)

This must occur at time:

t' = v_o sin(q)/  g

Then the maximum altitude is:

y _max =  - ½ g t' ²   + v_o sin(q) t' =

- ½ g [v_o sin(q)/  g] ²  +  v_o sin(q) [v_o sin(q)/  g]

=  v_o sin(q) ² /  2 g

Example Problem:

Find:

i) The maximum height attained when the launch angle for a projectile is q = 30 degrees.

ii) Write an equation for the slope (dy/dx)  of the path of the projectile at any point.

iii) Write the condition to obtain the range  R (maximum horizontal distance) of the projectile and find it.

iv) From (iv) what can be inferred about the angle F ?

Solutions:

i) The maximum altitude is:

y _max =  - ½ g t' ²   + v_o sin(q) t' =

- ½ g [v_o sin(q)/  g] ²  +  v_o sin(q) [v_o sin(q)/  g]

=  v_o sin(q) ² /  2 g    

q = 30 degrees:

=  v_o sin(30) ² /  2 g   

But sin (30) = 1/2 so:

y _max =  v_o (1/2) ² /  2 g   

=   v_o (1/4) /  2 g   =  v_o /  8 g

ii) Slope at any point: 

dy/dx =  (dy/dt)/ (dx/dt) =

 - ½ g t ² + v_o sin(q)/v_o cos(q)

iii) Range (R) occurs when y = 0  

 This occurs at double the time for max. altitude since trajectory is symmetric.

So: 2t'  =  2 ( v_o sin(q) / g ) 

Then x = R  =  v_o cos(q) 2 ( v_o sin(q) / g ) 

=  ( v_o ² / g) sin(2q)

iv)  The slope dy/dx is (expanded from (ii)):

dy/dx = -2 v_o sin(q) + v_o sin(q)/ v_o cos(q) 

= -  [sin(q)/ cos(q)] =   -  tan (q)

This shows the angle F =   p   -  q

i.e. projectile returns to Earth at same angle it left.

Suggested Problem:

Find the  x- and y-coordinates of the points on the trajectory of a missile launched at an angle of 80 degrees with an initial velocity of 100,000 f/s if the air resistance is 0.01mv. Find the value of x and y after 10 seconds.

Data:    q = 80^o,   k = 0.01mv   v = 10⁵ fs^-1 and t = 10s