pre calc 3.5 question

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 + C²) (g/2) (x/v)². in this formula, x is the horizontal distance in meters, y is the height in meters, v is the initial velocity in meters per second, g = 9.81 m/sec² is the acceleration due to gravity, and c > 0 is a constant determined by the angle of elevation. a howitzer fires an artillery round with a velocity of 884 m/sec.

a if the round must clear a hill 194 meters high at a distance of 2081 meters in front of the howitzer, what c values are permitted in the trajectory equation?

b if the goal in part a is to hit a target that n the ground 68 kilometers away is it possible to do so if so what values of C? If not what is the maximum distance the round will travel?

Answer 1

The constant C=tanθ ,,,,,θ is angle measured from x-axis

Comment back if you want to know why.

set in the equation y=195 m (1 m extra)

and x=2081

should have:

tan²θ-[(2*884²)/(9.81*2081)]tanθ+[(195*2*884²)(/9.81*2081²)]+1=0

tanθ=.11 or 76.53

yes, about can projectile reach 68km

set y=0 and x=68000

should have:

tan²θ-[(2*884²)/(9.81*68000)]tanθ+1=0

tanθ=1.17+/-.62