Question

why is the velocity in x direction becomes zero when a projectile hits an inclined perpendicularly

Answer 1

A body projected into the space and is no longer being propelled by fuel is called aprojectile.

Let’s say the incline plane is at some angle θ and I am throwing a ball on it. It’s going like that at some angle, at some other angle I have to choose right because θ is already taken. So let’s say the angle I throw it at is α. Then the ball will do something like that. Right? And go hit this somewhere. Now this is one way of looking at it but the same thing of throwing a projectile at an incline can be looked at like this also. What if I imagine that the ball is thrown? Right? Just like any other ball on the Earth. t so happens that it is an incline plane. Now this seems like a psychological difference right? Yeah it’s more to it than that. The way you solve the problem will differ. Why? Because there are basically two ways to solve this problem. Right? The problem of a ball on an incline plane. So in one you are frame of reference will be tilted like this. Yeah? Along parallel and perpendicular to the incline plane and the other you should say I will still keep it my old way, which is up and down, or sorry, right and left. Yeah? So overall you will have a horizontal and vertical itself. Now let’s take this one first r. The old one itself and see what happens. Let’s say you want to ask the question how long will this ball take to hit the incline plane. All you have now is the projectile that is going between the points, or rather passing through the points (x, x tanθ). What you are trying to find out is at what time t will its y-coordinate be x tanθ. Because you know who determines time? Only the vertical. So the advantage of this is more psychological because if you don’t like inclined planes in one step, I am giving you a way to remove the incline planes from your picture. From now the incline plane doesn’t matter at all. The incline plane only mattered to impose this constraint. That the y and x will be (x tanθ, x). You can solve the problem here. Let me show you. So now what matters to you? Only your y. Yeah? Only your y component matters to you for finding out the question of time. What’s happening along the horizontal line and the parallel direction here hardly matter. Because they don’t determine the time. The time is determined by the perpendicular. I am telling you all this to finally convince you psychologically that nothing is different between this diagram and this diagram except that you are replacing g with g cosθ. What does that mean for you? In this case you know what the answer for time of flight is. u sinθ to go up and u sinθ to come down. So overall 2u sinθ by g. Because that’s the only equation that we want right? So this is the equation that we want. Let’s take this with us and let’s see if our prediction made based on intuition is being matched over there. Okay? So the u sinθ, α into t part didn’t come with us. It chose not to so let’s write it down over here. u sinα into t. Yeah? We are sure this is the right equation right? Let’s go back and check. u sinα into t minus half g cosθ into t2. Yeah? I am being meticulous here because you can petrate again but in general these kind of things I only do in exams go wrong so I won’t do it now. So now let’s solve this it’s only going to take like a minute.