# Cartesian Convention of Displacement Vectors.

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1. Oct 26, 2014

### Hijaz Aslam

1. The problem statement, all variables and given/known data
Q.
A body dropped from a height H above the ground strikes an inclined plane at a height h above the ground. As a result of the impact, the velocity of the body becomes horizontal. The body will take the maximum time to reach the ground if :

(a) $h=\frac H4$ (b) $h=\frac H{2\sqrt{2}}$ (c) $h=\frac H2$ (d) $h=\frac H{\sqrt{2}}$

2. Relevant equations

$h=ut-\frac{gt}{2}$

3. The attempt at a solution
I've got the answer as $H/2$ and my text marks the answer valid.
By taking $t=t_1+t_2$. And taking $t_1=\sqrt{\frac{2(H-h)}{g}}$ and $t_2=\sqrt{\frac{2h}{g}}$. And taking
the differential coefficient of $t=\sqrt{\frac{2(H-h)}{g}}+\sqrt{\frac{2h}{g}}$ and equating it to 0, gives the value of $h=\frac{H}{2}$.

But I am confused about the direction of the entities involved in the motion. Here initially we take the origin as the point where we drop the body, taking 'up' as positive and 'down' as negative. But in the second case we consider the point where the body strikes the inclined plane as the origin and as in the former case 'up' as+ve and 'down' as -ve. But doesn't this variation in the origin in both the cases dismantle the final information about the motion, its time,height,velocity etc.

This doubt is related to my previous thread( https://www.physicsforums.com/threa...-not-touching-the-origin.778009/#post-4891774 ). Though I got a partially satisfactory answer for that question, the above question confuses me again.

2. Oct 28, 2014

### Vibhor

There is only one origin .In this problem you may consider it to be on ground .

I think you are using the equation $x = ut+\frac{1}{2}at^2$ .

The correct form is $x = x_0 + ut+\frac{1}{2}at^2$ ,where $x_0$ is the initial displacement. Use this with appropriate signs and you will get the correct answer .

Hope this helps .