Question

A body is dropped from a certain height and takes 6 seconds to reach the ground .how much time does it take to cover 1/9th the distance starting from top?​

Answer 1

Answer:

2 second

Explanation:

g=10 m/s^2

total height =h

time to cover h=6s

(v-u)/t=g

v=gt=10t

v^2=2gh=20h

100t^2=20h

3600=20h

h=180m

h2=180/9=20m

h2=1/2gt^2

(20*2)/10=t^2

t^2=4

t=2second

Answer 2

Given: Time, t = 6s

To Find: Time taken when the body covers 1/9th distance, t₂.

Solution:

To calculate T, the formula used:

• s = ut + (1/2)x at²
• here, s is the distance from where the body is dropped

        u is the initial velocity which is zero( u = 0 m/s)

        a is the acceleration due to gravity, g = 9.8 m/s²

        t is the time taken by the body to reach the ground

Applying the above formula:

s = 0 x 6 + (1/2) x 9.8 x 6²

 = 0 + (1/2) x 9.8 x 6 x6

 = 1/2 x 9.8 x 36

 = 1/2 x 352.8

 = 176.4 m

Now, as per the question, when the body covers 1/9th of s, the time taken by it, t₂ is calculated as:

s/ 9= ut₂ + (1/2)x g(t₂)²

176.4/9 = 0xt₂ + 1/2 (9.8 x t₂²)

176.4/9 = 0 + 1/2 (9.8 x t₂²)

176.4/9 = 1/2(9.8 x t₂²)

19.6 = 1/2 (9.8 x t₂²)

t₂² = 2x (19.6/ 9.8)

t₂² = 2 x 2

t₂² = 4

t₂ = √4

   = 2

t₂ = 2s

Hence, the time taken by the body to cover 1/9th of the distance is 2 seconds.