one body is dropped while the second body is thrown down words with initial velocity of 2 metre per second simultaneously the separation between them is 18 years after time
Hello,
The displacement of a body is given by s = ut + 1/2 * a * t^2
For a freely falling body, u = 0
Therefore, s1 = 1/2 * a * t^2 = 1/2 * 10 * t^2 = 5 * t^2
Therefore, s1 = 5 * t ^2
For a body falling with an initial velocity of 2 m/s, displacement after time t second will be, s2:
s2 = 2 * t + 1/2 *10*t^2
s2 = 2t + 5 * t^2
Now, when the separation between them is 18 m ( I have assumed 18 m in place of 18 years because I think you have typed it incorrect)
Therefore, s2 - s1 = 18
2t + 5 * t^2 - 5 * t^2 = 18
2t = 18
t = 9 seconds (Answer)
I guess the separation between them is 18 metres instead of 18 years. If it is 18 metres then answer of the question would be:-
let the required time be t.
For freely falling body we have,
S1=1/2gt²
=1/2×10×t²
=5t²
For body thrown downward with1m/s,
S2=ut+1/2gt²
=1×t+1/2×10×t²
=t+5t²
Now,S2-S1=18
t=18s
Hope it helps.
Good luck.
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