Two tanks are engaged in a training exercise on level ground. The first tank fir
ID: 2141569 • Letter: T
Question
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 240m/s at an angle 10.8? above the horizontal while advancing toward the second tank with a speed of 17.0m/s relative to the ground. The second tank is retreating at a speed of 35.0m/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.
A)Find the distance between the tanks when the round was first fired.
B)Find the distance between the tanks at the time of impact.
Explanation / Answer
let one is tank A and other is tank B,
the relative velocity of B with respect to A is,
vBA = vB - vA = 35 - 17= 18 m/s
imagine A at rest, B is moving away from with 18 m/s.
time of flight of shell
0 = uy.t - 0.5 gt^2
t = 2*uy/g
t = 2*u*sin(theta)/g
= 2*240*sin(10.8))/9.8
t= 9.1779 sec ...................................................................................................
horizontal distance traveled by the shell
x = ux*t
=240 cos(10.8)*9.1779
= 2163.68 m ....................................................................................................
This is the distance between the tanks at the time ofimpact .
at the starting,the separation is,
xo = x - vBA*t
if A is considered to be at rest then B moved forward with 18 m/s for t = 9.1779 sec
this additional distance gained,
vBA*t = 18*9.1779 = 165.20 m
Hence, the original distance when the round was first fired,
xo= x - vBA.t = 2163 - 165 = 1997.798 m ....................................................................
1)the distance between the tank when the round was first fired,is = 1997.798 m
2) the distance between the tanks at the time ofimpact .is = 2163.68 m
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