Question

tell me answer fast please​

Answer 1

it's your answer hopes it's help

Answer 2

Answer:

$v = \sqrt[10]{2} m {s}^{ - 1} \\ t = \sqrt{2} sec$

Step-by-step explanation:

If the ball is dropped gently , we consider initial velocity to be 0

$u = 0$

We have equations of motions which are valid here because of constant acceleration.

$2as = {v}^{2} - {u}^{2}$

Just putting values as in your case:-

$2 \times 10 \times 10 = {v}^{2} - {0}^{2}$

$200 = {v}^{2}$

$\sqrt[10]{2} = v$

Now using :-

$v = u + at$

$\sqrt[10]{2} = 0 + 10t$

$t = \sqrt{2} sec$

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