Math, asked by raj136kumarnirala, 8 months ago

tell me answer fast please​

Attachments:

Answers

Answered by mutharasanvlr
1

it's your answer hopes it's help

Attachments:
Answered by ankushsaini23
4

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

  • hope it helps you...
  • please mark it as a brainlist answer....
  • also please rate thanks and follow me...
  • stay home STAY SAFE....
Similar questions