two bodies begin to fall freely from the same height the second t second after the first.how long the first body begin to fall,will the distance between the two bodies be equal L.
To find the time at which the distance between the two falling bodies is equal to L, we need to determine the equations of motion for each body and solve for the time.
Let's assume that the initial height from which both bodies start falling is h. The first body starts falling at time t = 0 seconds, while the second body starts falling t seconds later.
For both bodies, we can use the equation of motion under constant acceleration:
h = ut + (1/2)at^2
Where:
- h is the height
- u is the initial velocity (0 since the bodies start from rest)
- t is the time
- a is the acceleration due to gravity (-9.8 m/s^2)
For the first body, we start from time t = 0:
h1 = (1/2)at^2
For the second body, we start from time t and subtract the additional time delay:
h2 = (1/2)a(t - t)^2
Simplifying h2, we have:
h2 = 0
Since the second body starts falling later and both have the same initial height, the equation becomes 0 = h1.
Now, we need to find the time when the distance between the two bodies is equal to L. Let's call this time T. We can subtract the equations of motion for both bodies:
h1 - h2 = L
(1/2)at^2 - 0 = L
Simplifying for t, we have:
t = √(2L/a)
Therefore, the first body will take √(2L/a) seconds to fall, and the distance between the two bodies will be equal to L at that time.