Thank you! I still have a doubt, what if the gun was shot downwards? Then the y component of the gun would be in the positive y direction? What would I do then? Thanks in advance.
Let the maximum height be h
work done by gravity during upward flight = -mgh (-as force is opposite to displacement, cos180 = -1)
work done by gravity during downward flight = mgh
Net force = -mgh+mgh = 0
I don't know the exact expression for air resistance, but since it always acts against...
Thanks!
I did consider the x and y components of the velocity of the center of mass separately.
As you said, there is no external force acting along the x axis, so I can apply conservation of linear momentum there.
0 + 0 = Mv1 + mucosθ
==> v1 = -(mucosθ)/M
So with that and my...
Work done is always force into displacement. As the object comes back to it's initial position, there is 0 net work on the body. So, I guess there is no work done by either force on the body. I may be wrong though, I'm not sure.
Hello everyone. First post here. I'll go the question straightaway.
Homework Statement
A gun of mass "M" placed on a smooth horizontal surface fires a bullet of mass "m" with a velocity "u" at an angle "θ" with the horizontal (ground). The velocity of the center of mass of the gun+bullet...