Recent content by Roul

So the lift force will be in the Y-direction, but the drag force will be at an angle (say theta) to the vertical direction, in between the horizontal flow velocity and the vertical particle velocity. So in my equation can I use FL and FDcos(theta) and rest will be the same, right?

In the attached picture, can I say that there is a lift force in the Y-direction, and a drag force too in the same Y-direction? FL proportional to V_fx2 ? FD proportional to V_py2 ? Is this equation of motion for the Y-direction correct here: ma = − mg − FD + V(rho)g + FL The lift force is...

Hi, thanks a lot for taking the time to respond in such details. From the equations, I get it that first I solve for the portion the force is acting and find out the velocity reached, and then use this velocity as initial velocity for the remaining trajectory. Thanks a lot. Ya I have no idea...

Hi, thanks so much. Ya, this makes so much sense. If I use finite delta(t) I will also need to use finite force, and the impulse concept will be disregarded a bit, I guess. Thanks a lot for pointing this out. Edit: However, when there is an impulse, there is an acceleration too. Isn't it. Or...

Ya, I wanted to know what delta(t) to use. Is there any standard that delta(t) should be such that the particle doesn't move by more than say 1mm, or 0.1*diameter, or any such thing. Is there any way to know what is the max. delta(t) allowed. When I use Impulse = Change in momentum = mv - 0...

Hi, thanks a lot for your reply and really sorry for my late response. Ya, I used mgh=0.5mv^2 to get the velocity, and used it in Impulse = mv-0 to get the impulse. Thanks.

Homework Statement [/B] Suppose I have a stationary object of mass 'm' and I want to apply a momentary force in the vertical direction so that it just reaches the height 'h'. So how do I calculate the impulse required in this case? Also how do I exactly define the delta(t) for the momentary...