Particle Motion in One Dimension

[vsc]

Two bodies begin a free fall from rest from the same height. If one starts 1.0 s after the other, how long after the first body begins to fall will the two bodies be 10 m apart?



x = x + vt +1/2at^2



I'm having trouble visualising a solution.

 

[ashishsinghal]

how did you get this equation:
x = x + vt +1/2at^2

This equation gives t = 0, t= -2v/a, none make any sense.

I will describe the situation.
The first ball would have traveled some distance at the time the second ball is thrown. Also the first ball would always have more speed than the second ball in their free fall situation. So the distance between them always increases. You need to find when it will be 10 m.

 

[Mr.A.Gibson]

Calculate the distance traveled and final velocity of the first ball after 1 second using the formula you stated.

x = x0 + vt +1/2at^2

Where x0 is the distance traveled by the first ball after one second. Now you are looking for a distance of 10m apart, so the distane traveled by the first ball x1 is 10m further on than the distance traveled by x2.
And since the initial speed and position of ball two is zero

x1 - x2 = x0 +vt +1/2at^2 - 1/2at^2

note the acceleration term canceled as both balls accelerate at the same rate.

10 = x0 + vt

Solve this for t and add on the first second when only one ball is falling.