Question

A stone is dropped from a certain height distance covered by it in one second

Answer 1

Stone is dropped therefore initial velocity will be zero

u=0

final velocity =…?

assume it x

and

time =1 sec

therefore

v=u + at (apply it)

put

v= gt (g= 10 m/s and u =0)

hence ,v=10×1=10 m/s

now to find height

v square = u square + 2as (Newton's third equation of motion)

v sq.= 2gs

100= 2×10×s

hence ,

s=5 m

therefore distance covered by it will be 5m

Answer 2

Answer:

5m

Explanation:

As initial velocity of ball is zero ..

time= t =1 sec

so apply Second equation of motion is

For body under gravity

h=vit+1/2gt²

h=0×1+1/2(10)(1)²

h=0+5(1)

h=5 m