Please help me friends... Give proper explanation... I will mark ur answer as a brainliest answer... A stone falling under Gravity. The total distance covered in the last second of its journey equal to the distance covered by it in the first three seconds of motion. The time for which the stone is in air is a) 5secb) 12secc) 15secd) 8sec. Created by nisha1456. Physics

Please help me friends... Give proper explanation... I will mark ur answer as a ...

AJAYMAHICH Distance covered by a body in the n th second is :

S(n) = u + 1/2*a*(2n - 1)

where, u = initial velocity, a = acceleration of the body

Let the stone remain in the air for t sec.

As, the stone falls from rest,

So, u = 0 m/s

Also, a = -g (acceleration due to gravity is directed downward) = -10 m/s^2

So, distance covered by the stone in the last second (or t th second) is :

S(t) = u + 1/2*a*(2t - 1)

= 0 - 1/2*10*(2t - 1)

= -5(2t - 1)

Distance covered by a body in n seconds is :

S = u*n + 1/2*a*n^2

So, distance covered by the stone in first 3 seconds is:

S = u*3 + 1/2*a*3^2

= 0*3 - 1/2*10*9

= 0 - 45

= - 45 m

Given,

S(t) = S

or, -5(2t - 1) = -45

or, 2t - 1 = 9

or, t = 5 sec

So, the stone remained in the air for 5 seconds.

Hope, it helps.

AJAYMAHICH wlcm dear dost

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Distance covered by a body in the n th second is :

S(n) = u + 1/2*a*(2n - 1)

where, u = initial velocity, a = acceleration of the body

Let the stone remain in the air for t sec.

As, the stone falls from rest,

So, u = 0 m/s

Also, a = -g (acceleration due to gravity is directed downward) = -10 m/s^2

So, distance covered by the stone in the last second (or t th second) is :

S(t) = u + 1/2*a*(2t - 1)

= 0 - 1/2*10*(2t - 1)

= -5(2t - 1)

Distance covered by a body in n seconds is :

S = u*n + 1/2*a*n^2

So, distance covered by the stone in first 3 seconds is:

S = u*3 + 1/2*a*3^2

= 0*3 - 1/2*10*9

= 0 - 45

= - 45 m

Given,

S(t) = S

or, -5(2t - 1) = -45

or, 2t - 1 = 9

or, t = 5 sec

So, the stone remained in the air for 5 seconds.

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