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.

Solution

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.

x-dimension

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.