Question

a body dropped freely from a certain height reaches the ground in 4s.the average velocity during its fall is (g=10m/s^2)​

Answer 1

Answer:

40 m/s

Explanation:

From the velocity-time relation,

         v=u+at

         v=0+gt

          v=gt

∴v=10×4

    =40 m/s

Answer 2

Concept:

Equations of motion

1) v = u + at

2) s = ut + 1/2at²

3) v² - u² = 2as

Given:

Time taken by the body to reach the ground, t = 4 s

Acceleration due to gravity, g = 10 m/s²

Find:

The average velocity during its fall.

Answer:

The average velocity of the body during its fall is 20 m/s.

Solution:

The given body is falling freely.

∴ Initial velocity, u = 0 m/s

Acceleration, a = g = 10 m/s²

Time taken, t = 4 s

Let the height from which the body is dropped be h.

Using the second equation of motion, we have

h = ut + 1/2at²

h = (0)(4) + 1/2(10)(4)²

h = 0 + 5(16)

h = 80 m

Total distance, h = Height = 80 m

Total time taken, t = 4 s

Now, average velocity is given by,

Average velocity, $v_{avg}$ = Total distance / Total time taken

                                    = h/t

                                    = 80/4

                                    = 20 m/s

Hence, the average velocity of the body during its fall is $v_{avg}$ = 20 m/s.

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