Question

A particle fals freely under gravity and covers the distance of 24.5 m in the final second of its motion . What is the total distance fallen by the particle?

Answer 1

Answer:

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Explanation:

Distance travelled in the last second(nth second)=24.5m

Assuming u=0 and g=10 m/s^2

According to the formula,

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

 24.5 = 0+10/2 (2n-1)

 24.5 = 5(2n-1)

 24.5 = 10n -5

  10n = 29.5

        n=2.95

        n~3 sec

That means, t=3 s

According to the second kinematical equation,

s = ut + 1/2 at^2

S = o +1/2 ×10× 9

S= 5 ×9  

S = 45m

Hence, answer to this question is 45m

(Note: this is an approximate answer. Hence you can choose the option closest to this in the examinations. Remember, approximation is important to save time and solve sums with mininmum errors and hence save time!)

Hope this helped:)