A ball is allowed to fall freely from certain height. What distance it

A ball is allowed to fall freely from certain height. What distance it covers in 1st second?

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

g/2

d = 0.5g*t^2 = 4.9*1^2 = 4.9 m.

g/2

thanks for the answer ayehsa and hoorub also both answers were correct.

0.5×g×t^2

0.5=1\2 & t=1
So
1/2×g×(1^2)
So,
g/2
4.9m

g/2

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.

To determine the distance the ball covers in the first second, we can make use of the equations of motion. Specifically, we can use the equation for the distance covered during free fall, assuming no air resistance:

s = u * t + (1/2) * g * t^2

Where:
- s is the distance covered
- u is the initial velocity (which is 0 since the ball starts from rest)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time

Since the ball is allowed to fall freely, it starts from rest, meaning the initial velocity (u) is 0 m/s. Plugging in this value, along with t = 1 second and g = 9.8 m/s^2 into the equation, we can calculate the distance covered in the first second. Let's do the math:

s = 0 * 1 + (1/2) * 9.8 * (1^2)
s = 0 + (1/2) * 9.8 * 1
s = 0 + 4.9
s = 4.9 meters

Therefore, the ball will cover a distance of 4.9 meters in the first second of free fall.

4.9m