Mr. Rogers' AP Physics C: Projectile motion Problems



Anti-Balloon Gun Problem

An observation balloon floats over the Enemy lines and the defenders decide to shoot it down. The defending gunners determine the balloon is 3015 m above the ground at a distance of 5000 m. They fire a cannon at an angle of 40 with respect to the horizon with an initial velocity of 760 m/s.



Place the origin at the bottom of the anti-balloon gun. 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 x-dimension 

vox  =  vo cos q

        = (760 m/s) cos 40

        = 585 m/s


x = 1/2axt2 + voxt         

      = 0 + voxt

Solving for time gives:

t = x / vox

     = (5000 m) / (585 m/s)

t  = 8.55 sec

Finish in y-dimension >>>

y- dimension














voy  =  vo sin q

        = (760 m/s) sin 40

        = 488 m/s

y = 1/2ayt2 + voyt

  = 1/2 (-9.8 m/s2)(8.55 sec)2 + 488 m/s)(8.55 sec)

y = 3814 m

While the actual height is 3815 m the cannon shell will destroy the balloon if the fuse is set so that it explodes 3814 m off the ground.


Conclusion and Significance

Typically the gun crew would be able to measure the range and height, but they would have to calculate the angle. Air resistance and wind would complicate the situation.