A ball is allowed to freely from certain height.It covers a distance in 1st second equals ti:
A)2g B)g C) g/2 C)none
h = 0 + 0.5(g)(1)^2 = 0.5g
So, option (C)
g/2
The distance covered by a ball in the first second can be calculated using the formula for the distance traveled by an object in free fall:
Distance = Initial velocity * time + (1/2) * acceleration * time^2
In this case, the ball is dropped from rest, so the initial velocity is 0 m/s. The acceleration due to gravity is represented by "g" (approximately equal to 9.8 m/s^2).
Substituting these values into the formula:
Distance = 0 * 1 + (1/2) * g * 1^2
Distance = 0 + (1/2) * g * 1^2
Distance = 0 + (1/2) * g * 1
Distance = (1/2) * g
Therefore, the correct answer is C) g/2.
To determine the answer to this question, we need to understand the concept of motion under gravity.
When an object is freely falling under the influence of gravity, it experiences a constant acceleration, which is denoted by 'g.' The value of 'g' is approximately 9.8 m/s² on the surface of the Earth.
The equation that relates the distance (s) covered by a freely falling object in time (t) is given by:
s = ut + (1/2)gt²
Where:
s = distance
u = initial velocity (which is zero for a freely falling object)
g = acceleration due to gravity
t = time
In this case, we are interested in the distance covered by the ball specifically in the first second.
Let's substitute the given values into the equation:
s = 0 * 1 + (1/2) * g * (1^2)
s = 0 + (1/2) * g
s = g/2
Therefore, the distance covered by the ball in the first second is equal to (g/2).
So, the correct answer is C) g/2.