a body is released from a height and falls freely towards the earth. exactly one sec later another body is released.what is the distance between the two bodies 2sec after the release of the second body. if g=9.8m/s^2.
hf1 is the final height of first
hf2 is the final height of second.
so,the question is
hf1-hf2=ho1-4.9t^2-ho2+4.9(t-1)^2
t=2; ho1=ho2
To find the distance between the two bodies 2 seconds after the release of the second body, we can use the equation of motion for free-falling objects. In this case, the equation can be written as:
d = 1/2 * g * t^2
Where:
- d is the distance traveled
- g is the acceleration due to gravity (given as 9.8 m/s^2)
- t is the time
For the first body, which is released 1 second earlier, the time (t1) would be 2 seconds. Let's calculate the distance traveled by this body:
d1 = 1/2 * g * t1^2
= 1/2 * 9.8 * (2)^2
= 1/2 * 9.8 * 4
= 19.6 meters
Now, for the second body, which is released 1 second after the first body, the time (t2) would be 1 second. We want to find the distance traveled by this body 2 seconds after its release. So, the total time for the second body would be 2 seconds.
d2 = 1/2 * g * t2^2
= 1/2 * 9.8 * (2)^2
= 1/2 * 9.8 * 4
= 19.6 meters
Therefore, the distance between the two bodies 2 seconds after the release of the second body is equal to the distance traveled by each body, which is 19.6 meters.