please solve the above question whose answers are 50m and 25m respectively
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If the bodies are dropped simultaneously, the initial separation between them will always remain constant until one of them reaches the ground. This is because, both are released at same time AND have same acceleration downwards, and using this we can say that relative to one body, the other is always at rest, that is, until it reaches the ground. So, lets first check when the first body will touch the ground. Obviously, the one at 100m will reach first.
[tex] \frac{1}{2}g t^{2} = h 0.5*10* t^{2} = 100 t^{2} = 20 t approximately is between 4 and 5. [/tex]
So, at t=3, the distance will be constant and that is 50 m.
But now at t=5 sec, the first body will have reached the ground, so the distance between them is now the distance between the second body and the ground. So lets calculate distance travelled by second body in 5 sec.
[tex] \frac{1}{2}g t^{2}=h 0.5*10*25=h h=125m [/tex]
So, the distance travelled by the second body is 125. Initial height was 100m, so the distance from ground is 25 m and is consequently distance from the first body.
[tex] \frac{1}{2}g t^{2} = h 0.5*10* t^{2} = 100 t^{2} = 20 t approximately is between 4 and 5. [/tex]
So, at t=3, the distance will be constant and that is 50 m.
But now at t=5 sec, the first body will have reached the ground, so the distance between them is now the distance between the second body and the ground. So lets calculate distance travelled by second body in 5 sec.
[tex] \frac{1}{2}g t^{2}=h 0.5*10*25=h h=125m [/tex]
So, the distance travelled by the second body is 125. Initial height was 100m, so the distance from ground is 25 m and is consequently distance from the first body.
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