two body begins a free fall from rest from the same hight 1sec apart how long after the first body begins to fall when the two bodies be 10m apart
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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 .
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 .
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