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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.
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x-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
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 |
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.