The distance covered by a freely falling body can be calculated using the equation of motion:

s = ut + 0.5gt^2

where:
s = distance covered
u = initial velocity
t = time
g = acceleration due to gravity

In this case, the body is freely falling, so the initial velocity (u) is 0. The acceleration due to gravity (g) is given as 10 m/s^2, and the time (t) is 3 seconds.

Substituting these values into the equation gives:

s = 0*3 + 0.5*10*3^2
s = 0 + 0.5*10*9
s = 0 + 45
s = 45 meters

So, the distance covered by the body during the first three seconds of its motion is 45 meters.

Question

The distance covered by a freely falling body can be calculated using the equation of motion:

s = ut + 0.5gt^2

where:
s = distance covered
u = initial velocity
t = time
g = acceleration due to gravity

In this case, the body is freely falling, so the initial velocity (u) is 0. The acceleration due to gravity (g) is given as 10 m/s^2, and the time (t) is 3 seconds.

Substituting these values into the equation gives:

s = 0*3 + 0.5*10*3^2
s = 0 + 0.5*10*9
s = 0 + 45
s = 45 meters

So, the distance covered by the body during the first three seconds of its motion is 45 meters.

Knowee AI · Accepted Answer

The distance covered by a freely falling body can be calculated using the equation of motion:

s = ut + 0.5gt^2

where:
s = distance covered
u = initial velocity
t = time
g = acceleration due to gravity

In this case, the body is freely falling, so the initial velocity (u) is 0. The acceleration due to gravity (g) is given as 10 m/s^2, and the time (t) is 3 seconds.

Substituting these values into the equation gives:

s = 0*3 + 0.5*10*3^2
s = 0 + 0.5*10*9
s = 0 + 45
s = 45 meters

So, the distance covered by the body during the first three seconds of its motion is 45 meters.

What is the distance covered by a freely falling body during the first three seconds of itsmotion? (g=10m/s2

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