The situation can be modeled as though the 1st tank were stationary, and the 2nd tank were retreating at 20 m/s.
The horizontal and vertical components of the shell's speed at firing are
x: 43.4 m/s
y: 246.2 m/s
So, the height of the shell is
h(t) = 246.2t - 4.9t^2
It hits the ground at t=50.24 seconds
During that time, the shell travels horizontally 50.24*43.4 = 2180.4 meters
The second tank recedes 20*50.24 = 1004.8 meters
For the shell to hit the 2nd tank, the tank must have initially been 2180.4-1004.8 = 1175.6 meters ahead.
Any of you physics whizzes have a less cumbersome solution?
Two tanks are engaged in a training exercise on level ground. The first tank fires a paint-filled training round with a muzzle speed of 250 m/s at 10.0◦ above the horizontal while advancing toward the second tank with a speed of 15.0 m/s relative to the ground. The second tank is retreating at 35.0 m/s relative to the ground, but is hit by the shell. You can ignore air resistance and assume the shell hits at the same height above ground from which it was fired. Find the distance between the tanks
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