Find the distance between the tanks when the round was first fired? Find the dis

Question

Find the distance between the tanks when the round was first fired?
Find the distance between the tanks at the time of impact.
Alternative Exercise 3.102 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 239 m/s at an angle 10 8 o above the horizontal while advancing toward the second tank with a speed of 16.0 m/s relative to the ground. The second tank is retreating at a speed of 39.0 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

Explanation / Answer

solve for B first by tying your reference frame to the shooting tank

Initial vertical velocity
vy = 239sin10.8
vy = 44.784 m/s

which becomes zero in the gravity field in

t = 44.784 / 9.81
t = 4.565 seconds

total flight time is
2t = (2)4.565
2t = 9.13.s

horizontal velocity is
vx = 239cos10.8
vx = 234.7666. m/s

for a distance between them at impact of

d = vx(2t)
d = 234.7666(9.13)
d = 2143.49  <==ANS (B)

the velocity difference between the tanks is

v = 39 - 16
v = 23 m/s

the change in distance between the tanks while the shell flew is

d = 23(9.13)
d = 209.997
d = 210 m

so the distance between the tanks when the round was fired is

d = 2143.48997 - 209.9969
d = 1933.49 <===ANS (A)

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