Trajectory Projectile

Question

A projectile fired from the point (0,0) at an angle to the positive x-axis has a trajectory given by y=Cx-(1+C^2)(g/2)(x/v)^2. A howitzer fires an artillery round with a velocity of 876m/sec. If the round must clear a hill 232 meters high at a distance of 2490 meters in front of the howitzer, what C values are permitted in the trajectory equation?

Answer 1

What information does this question give us?

The Howitzer shoots the round at (0,0)
v = 876 m/sec (initial velocity)
x = 2490 m (distance along x-axis we are interested in)
y = 232 m (minimum height to clear the hill)
g = 9.81 m/s²  (acceleration due to gravity) not sure why the value of g was not stated in the question.
C = (angle from x-axis to shoot from) I assume to be some angle in degrees

The question asks for two values, we can see this is because there is a value squared so we can use the quadratic equation. First rewrite the equation so the quadratic equation can be used. The unknown variable is C.

Expand the original equation then reorganize to something like this:

 

0 = -(g/2*(x/v)^2+y) + (x)C + (-g/2*(x/v)^2)C^2

0 = c + bC + aC^2

 

Note: The quadratic formula is usually set up like this ax² + bx + c = 0 where x is C in your case. You may want to reorganize the above equation to reflect that.

 

Substituting the values in for the variables will simplify the equation:

c =  -(g/2*(x/v)^2+y) = -271.62 m

b = x = 2490 m

a = (-g/2*(x/v)^2) = -39.617 m

 

C = (-b ± √(b^2 - 4*a*c)) / (2*a) (quadratic equation)

C = (-2490 m ± √((-271.62 m)^2 - 4*(-39.617 m)*(-271.62 m))) / (2*(-39.617 m))

 

C = 0.11^o and 62.74^o

 

Hope that helps Jessica.