Two tanks are engaged in a training exercise on level ground. The first tank fir

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 246 at an angle 10.0 above the horizontal while advancing toward the second tank with a speed of 12.5 relative to the ground. The second tank is retreating at a speed of 31.0 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 when the round was first fired.
Take free fall acceleration to be = 9.80 .

Explanation / Answer

in this problem there is no mentioned units, but take all units in S.I units
velocity of the first tank is u =246 m/s angle with horizontal =100 first tank advancing toward the second tank with a speed of v =12.5 m/s ______________________________________________________________________________ we consider the motion of first tank as a projectile the time taken for first tank hit to second tank is t =2usin/g                                                                             =2*246*sin(100)/9.8                                                                             =8.7 s the distance travelled by first tank before hit second tank is D =ucos*t   (horizontal case acceleration is zero s =ut+1/2at^2) D =246*cos100*8.7     =2107.68 m the distance between the tanks when the round was first fired is d =D -v*t                                                                                                    =2107.68 -12.5*8.7                                                                                                    =1998.93 m first tank advancing toward the second tank with a speed of v =12.5 m/s ______________________________________________________________________________ we consider the motion of first tank as a projectile the time taken for first tank hit to second tank is t =2usin/g                                                                             =2*246*sin(100)/9.8                                                                             =8.7 s the distance travelled by first tank before hit second tank is D =ucos*t   (horizontal case acceleration is zero s =ut+1/2at^2) D =246*cos100*8.7     =2107.68 m the distance between the tanks when the round was first fired is d =D -v*t                                                                                                    =2107.68 -12.5*8.7                                                                                                    =1998.93 m

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