Question

A ball is free fall from a height of 500 m. Then find time at which ball reached the ground and velocity at height of 250 m ​

Answer 1

please refer to the attachment given....hope this will helpful to you dear

Answer 2

Answer:

Velocity at that point is 50√2 m / s.

Explanation:

At the initial position : Velocity( u ) of ball was 0.

Using the equations of motion :

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

= > Total height = initial velocity x time taken + half of acceleration x time^2

= > 500 m = 0 x time taken + 1 / 2 ( a x time taken^2 )

Here, body is falling under gravity, due to gravitational force, with the acceleration g ( acceleration due to gravity, it's value is assumed as constant ( here, 10 m/s^2 )

= > 500 m = 1 / 2 ( 10 x time^2 ) m / s^2 x s^2

= > 500 x 2 / 10 = time^2

= > 100 = time taken^2

= > 10 = time taken

= > 10 sec = time taken ( in seconds )

At the height of 250 m from that point :

Using v^2 = u^2 + 2aS { where symbols have their usual meaning }

= > velocity at that point ^2 = 0 + 2( g )( 250 ) m / s^2 x m

= > velocity at that point ^2 = 500g ( m / s )^2

= > velocity at that point = 50√2 m / s