Question

A body of mass 5 kg starts from rest and falls freely. The distance covered by it in one
second is
A.9.8 m
B.980 m
C.49 m
D.4.9 m​

Answer 1

Answer:

Use s = (g/2)(2n - 1) and put n = 1 ....because initial velocity u = 0

that should do it !

Explanation:

plz like

Answer 2

Explanation:

GIVEN; MASS OF BODY =>. 5 KG

g ( ACCELERATION DUE TO GRAVITY) = 9.8 M/S^2

TIME OF FREE FALL FOR BODY = 1 SECOND

INITIAL VELOCITY = 0 M/S

DISTANCE TRAVELLED = S

SOLUTION;

AS WE DON'T KNOW FINAL VELOCITY

BY USING SECOND EQUATION OF MOTION ,WE GET ;

S = UT + 1/2 AT^2

S = (0)(1) + 1/2 (9.8)(1)^2

S = 0 + 4.9 × 1

S = 4.9 M

SO , DISTANCE TRAVELLED BY BODY OF MASS 5 KG IN 1 SECOND IS 4.9 M

THANKS FRIEND, HOPE IT HELPS .