Physics, asked by nandnitharani3, 3 months ago

a body is falling freely under gravity how much distance it falls during an interval of time between 2nd and 3rd seconds of its motion
a 5m b 15m c 20m d 25m​

Answers

Answered by abhi569
10

Answer:

25 m

Explanation:

Between 2nd and 3rd seconds means 'in 3rd second'.

Using, Sₙ = u + a/2 (2n - 1)

In question,  

u = 0(at rest), a = g = 10 , n = 3

 Therefore,

S₃ = 0 + (10/2) [2(3) - 1]

    = (5) (6 - 1)

    = 25 m

Answered by Anonymous
135

Given :-

  • A body is falling freely under gravity.

To Find :-

  • How much distance it falls during an interval of time between 2nd and 3rd seconds of its motion.

Formula Used :-

 \longmapsto \sf\boxed{\bold{\pink{S_n =\: u + \dfrac{a}{2}(2n - 1)}}}

where,

  • n = Time
  • u = Initial velocity
  • a = Acceleration

Solution :-

Given :

  • Initial velocity (u) = 0 m/s
  • Acceleration = 10 m/s²
  • Time (n) = 3 seconds

According to the question by using the formula we get,

 \implies \sf S_3 =\: 0 + \dfrac{\cancel{10}}{\cancel{2}}(2 \times 3 - 1)

 \implies \sf S_3 =\: 0 + 5(6 - 1)

 \implies \sf S_3 =\: 5(5)

 \implies \sf S_3 =\: 5 \times 5

 \implies \sf\bold{\red{S_3 =\: 25\: m}}

\therefore The distance it falls during an interval of time between 2nd and 3rd seconds of its motion is 25 m .

Hence, the correct options is option no (d) 25 m .

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