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A body is falling freely from certian height. If thebody coves 36% of total height in the last secondof its motion, then the height from which it is fallingis(1) 98 m(3) 44.1 m(2) 122.5 m(4) 149 m​

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Answer:

Explanation:

TOTAL distance travelled = h

total TIME taken = n seconds

distance travelled in the last(n th) second = 0.36h

We know that distance travelled in nth second = u + (a/2) (2n-1)

⇒ 0.36h = 0 + (9.8/2) (2n-1)

⇒ 0.36h = 9.8n - 4.9   --------------------------------(1)

We know that h = ut + (1/2) at²

⇒ h = 0×t + (1/2) × 9.8 × n²

⇒ h = 4.9n²  ---------------------------------------------(2)

Replacing VALUE of h from (2) in (1),

0.36 × 4.9n² = 9.8n - 4.9

Divide LHS and RHS by 4.9, we get

⇒ 0.36n² = 2n - 1

⇒ 0.36n² - 2n + 1 = 0

n= \frac{-b \pm \sqrt{b^2-4ac}}{2a}\\ \\n= \frac{2 \pm \sqrt{4-4 \times 0.36 \times 1}}{2 \times 0.36}\\ \\ n= \frac{2 \pm \sqrt{2.56}}{0.72}\\ \\n= \frac{2 \pm 1.6}{0.72}\\ \\n= \frac{2+1.6}{0.72} \ \ or\ \ \frac{2-1.6}{0.72}\\ \\n= \frac{3.6}{0.72} \ \ or\ \ \frac{0.4}{0.72}\\ \\ n = 5s \ \ (\text{ignoring the other value})

h = 4.9n² = 4.9 × 5² = 122.5m

Answer is 122.5m


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