Question

two tanks are 2.0*10^3 m apart and moving towards each other. One tank is moving at 10 m/s and fires a shell at the other one at an angle of 15 degrees. If the other tank is moving at 40 m/s, at what velocity, neglecting air resistance, would the shell jave to be fired at to hit the other tank?

Answer 1

By finding the relative velocity of the shell to the 2nd tank, we can simplify this problem to mere projectile motion:
The velocity of the shell relative to the ground is it's x component plus the speed of the firing tank:
\[u_(shell,ground) = u*\cos(15) + 10\]
thus, the velocity of the shell relative to the 2nd tank is:
\[u_(shell,2nd tank) = u(shell,ground) - u(2nd tank,ground) = u*\cos(15) + 10 - (-40) = u*\cos(15) + 50 \]
Now we simply consider a projectile moving in the x and y directions with the velocities:
\[u_x = u*\cos(15) + 50\]
\[u_y = u*\sin(15)\]
From the y component we calculate the time it takes for the projectile to fall:
\[0 = u_y - 0.5*g*t^2\]
dividing by t (\[t \neq 0\] ) we get:
\[t = \frac{ 2*u*\sin(15) }{ g }\]
And finally, we substitute the above time into the constant-velocity x equation:
\[x = u_x*t\]
\[2*10^3 = (u*\cos(15) + 50)*\frac{ 2*u_1*\sin(15) }{ g }\]
and after a few simplifications we solve the quadratic equation:
\[u_(1,2) = \frac{ -2.6 \pm 20.17 }{ 0.1 }\]
and, leaving out the negative result, get:
\[u = 175.7 (m/\sec)\]

Answer 2

seems the 2nd equation is not visible. Here is the missing part:
\[u(shell,2ndtank) = u*\cos(15) + 10 - (-40) = u*\cos(15) + 50\]