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Let's say we have a ball and we dropped it from height $20 \;\text{m}$

Case 1: Taking downward direction as positive
$$v^2 = u^2+2as$$ $$ v^2 = 0 + (2)(10)(20) $$ $$ v = 20 \;\text{m/s}$$ Nothing wrong until now but check this out.

Case 2: Taking upward direction as positive
$$(-v^2) = u^2+2(-10)(-20)$$ $$v^2 = 0 +400$$ $$v = +20 \;\text{m/s}$$

How can $v$ be positive when we assumed downward direction as negative? Am I misinterpreting something wrong?

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  • $\begingroup$ $(-20)^2=400$ as well, :) $\endgroup$
    – Physiker
    Commented Jun 27, 2021 at 10:25
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    $\begingroup$ but mathematically it should be negative as final velocity is downward direction and we assumed it as negative? $\endgroup$
    – Rambal heart remo
    Commented Jun 27, 2021 at 10:28
  • $\begingroup$ Yes. Read the answer below. $\endgroup$
    – Physiker
    Commented Jun 27, 2021 at 10:33

1 Answer 1

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The ball could have been thrown upwards from the ground with a velocity of 20 m/s upwards, reached the peak of its trajectory at a height of 20 m, then hit the ground again with a speed of 20 m/s downwards. The equations of motion know nothing about the history of the ball. That is why $v^2=u^2+2as$ will give two possible values for $v$, one positive and one negative. You have to choose the relevant value from the context of the question.

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