Bomber Problem

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 >>>

v_ox = v_ocos q

= (760 m/s) cos 30°

= 658 m/s

D_x = 1/2a_xt² + v_oxt

= 0 + v_oxt

= (658 m/s) (77.6 sec.)

= 51, 060 m

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

v_oy = v_osin q

= (760 m/s) sin 30°

= 380 m/s

y = 1/2a_yt² + v_oyt

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

y = 0

0 = 1/2a_yt² + v_oyt

0 = 1/2a_yt + v_oy

- 1/2a_yt = v_oy

t = -(2 / a_y) (v_oy)

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.