Question

Q#an object falls from rest from a height of 10 meter above the ground surface, find time to reach earth(1) suface(2) speed at time of hitting earth. Assume earth acceleration due to gravity. a=10m/s^2.​

Answer 1

Answer:

√2 s     ;    10√2 ms⁻¹

Explanation:

  Here, given :

    Object falls( & not thrown ), it means that was initially at rest where it's velocity was 0.

 This means,

   Initial velocity = u = 0

Let the velocity at which it hits the ground be v( final velocity ).

  Also, height is 10m , S = 10m and acceleration acting is g.

  Using S = ut + (1/2)at^2    { symbols have their usual meaning }

⇒ 10 = ( 0 )t + (1/2)(g)t^2

⇒ 10 = (1/2)10.t^2

⇒ 10 x 2 / 10 = t^2

⇒ 2 = t^2

⇒ √2 = t

(1)  Time is √2 sec or  1.414 sec.

 Using v = u + at;  

⇒ v = 0 + (g)√2

⇒ v = g(√2)

⇒ v = 10√2

  Velocity just before hitting the ground is 10√2 ms⁻¹.