Question

two objects begin a free fall from rest from the same height 1.00 s apart. how long after the first object begins to fall will the two objects be 10.0 m apart?

Answer 1

Assumption:

We assume that the vectors pointing downwards are taken to be positive, and since both the objects were released from rest, their initial speed is zero.

Given:

Initial speed of first object, v01 = 0 m/s.

Initial speed of second object, v02 = 0 m/s.

Acceleration due to gravity, g = 9.8 m/s2.

Distance of separation between the first and second object,x = 10.0 m.

232-1382_1.PNG

Since the object falls freely under the action of gravity, it accelerates the object with magnitude corresponding to the acceleration due to gravity that is a = g .

Also, the object was at the location of observer at t = 0, therefore x01 = 0 m .

Substitute the above conditions to have

232-38_2.PNG

Since the second object was released one second later, the distance travelled by the object corresponds to time t-1 .

Therefore the distance (say x2) travelled by the second object, localized at the very same moment, is given as:

232-1872_3.PNG

Since the object falls freely under the action of gravity, it accelerates the object with magnitude corresponding to the acceleration due to gravity that is a = g.

232-1075_4.PNG

The distance of separation (say x) between the two objects is given as the difference of distance travelled by the first object ad the distance travelled by the second object, that is:

x = x1 – x2

Substitute the value of x1 and x2 from equation (1) and (2) in the relation above

232-140_5.PNG

To avoid confusion, we write dimensions along with the variables:

232-113_6.PNG

Now our equation is dimensionally correct and hence the above equation can be written as:

232-1821_7.PNG

Substitute the given value of x in the equation above to have

232-1111_8.PNG

As we assumed above that the dimensions of t is [s], it is clear that the time after which the objects would be separated by 10 m is 1.52 s .