# Question #b7271

Jan 12, 2018

If the bus is moving with constant velocity then you will see that the coin will fall directly downwards in a linear pathway,because you are moving with the same speed with which the coin will also move forward and drop down due to gravity.
In this case relative velocity of coin horizontally is (V coin - V observer)=0
If you are moving with an acceleration then you will see the coin going down with a relative velocity backwards in a parabolic pathway as when you release the coin it will have the same velocity to yours but in the next second your velocity will increase due to acceleration but horizontally the coin will have no acceleration hence that constant velocity with which you released it,the coin will be moving
Here,relative velocity of coin horizontally (V coin - V observer) is negative in value as the later is more.
Now if you are going in retardation you will see it going down in front of you along a parabolic pathway.
As here,
relative velocity=(V coin - V observer) is positive as your velocity after releasing the coin has decreased w.r.t what was earlier due to retardation.
Now,suppose when you are moving with acceleration and released the coin to see its pathway,its position can be defined as
X1=movement horizontally=$\left(u \cdot t\right)$...1
Where,u is the relative velocity and X1 displacement occurs in time t
In that same time t vertically if it goes X2 distance,then we can write,
X2= $\left(\frac{1}{2}\right) g {\left(t\right)}^{2}$...2
Eliminating t from 1&2 we get,
X1= $\frac{2 \cdot g \cdot {\left(X 2\right)}^{2}}{{u}^{2}}$
Which is an equation of parabola