Mr. Rogers' AP Physics C: Projectile motion Problems



Artillery's Range Problem

An  artillery shell is fired at an angle of 30 with respect to the horizon with an initial velocity of 760 m/s. Find the range of the artillery shell.



Place the origin at the bottom of the cannon. Define horizontal as the x-dimension with negative to the left and vertical as the y-dimension with negative downward.


Start by finding time
in  the y-dimension  >>>
















vox  =  vo cos q

        = (760 m/s) cos 30

        = 658 m/s


Dx = 1/2axt2 + voxt         

      = 0 + voxt

      = (658 m/s) (77.6 sec.)

      = 51, 060 m

    x  = 31.7 miles

x  = 31.7 miles

Note: In this problem the projectile is going faster than the speed of sound. Under such conditions, a real artillery shell's range would be significantly less less due to air resistance.

y- dimension


voy  =  vo sin q

        = (760 m/s) sin 30

        = 380 m/s


y = 1/2ayt2 + voyt

but an instant before the shell hits, it will be zero meters above the ground, hence:

y = 0

0   = 1/2ayt2 + voyt

0   = 1/2ayt + voy

- 1/2ayt =  voy

t  =  -(2 / ay) (voy)

t  =  -(2 / (- 9.8 m/s2)) (380 m/s)

   = 77.6 sec


<<<   Finish in x-dimension

Conclusion and Significance

Typically, artillery crews would be able to measure the range to the target with a range finder. They would have to find the correct angle. these physics are more complex.


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