Question

a stone dropped freely from 72 m height of the tower and reaches to the ground in 6 seconds. calculate the acceleration due to gravity of that stone​

Answer 1

The acceleration due to gravity of that stone is 4 $m/s^2$.

Given,

The height of the tower, h = 72 m

The time taken to reach the ground, t = 6 seconds

To find,

The acceleration due to the gravity of that stone, g.

Solution,

Let the initial velocity u = 0

We can easily solve this question using the equation of motion,

S = ut + 1/2 a$t^2$

where S = displacement,

u = initial velocity,

a = acceleration, and

t = time taken

In this case, the height will be the displacement, so

72 = 0×t + 1/2 g$t^2$

⇒ g = (72 × 2) / 36

⇒ g = 4 $m/s^2$

So, in the given situation the acceleration due to gravity is 4 $m/s^2$.

Answer 2

Answer:

4 m/sec²

Explanation:

Given : height = 72 m

            time = 6 sec

as, the stone is dropped freely its initial velocity will be zero.

     by using equation of motion

      s = ut + 1/2 at²

s = distance

u = initial velocity

t= time

a = acceleration

but in this case as gravity is acting the equation will be

s = ut + 1/2 gt²

72 = 0[6] + 1/2 g[6]²

72 = 0 + 1/2 g [36]

g = 72*2/36

g = 4 m/sec²

Therefore, the acceleration due to gravity will be 4 m/sec²

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