Asked by Ali

A ball is allowed to fall freely from certain height. What distance it covers in 1st second?
Answered by Henry
d = 0.5g*t^2 = 4.9*1^2 = 4.9 m.


Answered by Anonymous
As we know that
S=Vit+½at^2. a=g
OR
S=Vit+(0.5)gt^2
As Vi=0 then it becomes
S=(0.5)gt^2. g=9.8m/s..... (1)
Putting values in eq (1).
S=(0.5)(9.8)(1^2)
S=(0.5)(9.8) 0.5=½
OR
S=g/2
S=4.9m
Answered by Ayesha
g/2
Answered by Hoorub
g/2
Answered by fahad
thanks for the answer ayehsa and hoorub also both answers were correct.
Answered by Madiha
0.5×g×t^2
0.5=1\2 & t=1
So
1/2×g×(1^2)
So,
g/2
4.9m
Answered by Zain
4.9m
Answered by Malik umair
g/2
Answered by Bot
The distance covered by the ball in the first second is:

d = 0.5*g*t^2 = 0.5*9.8*(1^2) = 4.9 meters.

Therefore, the ball covers 4.9 meters in the first second of its free fall.