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Natalija [7]
3 years ago
11

Ball is dropped from the height H the total distance covered in last second of its motion is equal to distance covered in first

3 second. find the height.
pls answer this question correctly and don't give any irrelevant or else I will report the answer.​...
Physics
1 answer:
Alik [6]3 years ago
7 0

\rm{\gray{\underline{\underline{\blue{GIVEN:-}}}}}

  • A ball is dropped from a height = H

  • The total distance covered in last second of its motion is equal to the distance covered in first 3 second .

\rm{\gray{\underline{\underline{\blue{TO\:FIND:-}}}}}

  • The height of the journey .

\rm{\gray{\underline{\underline{\blue{SOLUTION:-}}}}}

We have know that,

\green\bigstar\:\rm{\gray{\overbrace{\underbrace{\red{S\:=\:ut\:+\:\dfrac{1}{2}\:at^2\:}}}}}

Where,

  • S = Distance .

  • u = initial velocity = 0m/s

  • t = time = 3s

  • a = acceleration

[Note :- Here, acceleration is ‛acceleration due to gravity’ .]

  • a = 10m/s^2

=> S = 1/2 × 10 × (3)^2

=> S = 5 × 9

=> S = <u>4</u><u>5</u><u>m</u> -----(1)

✒ If the ball takes ‛n’ second to fall the ground, then distance covered in nth second is,

\green\bigstar\:\rm{\gray{\overbrace{\underbrace{\red{S_n\:=\:u\:+\:\dfrac{g}{2}\:(2n\:-\:1)\:}}}}}

=> Sn = 0 + 10/2 (2 × n - 1)

=> Sn = 5 (2n - 1)

=> Sn = 10n - 5 -----(2)

Therefore,

  • 10n - 5 = 45

=> 10n = 45 + 5

=> n = 50/10

=> n = <u>5</u>

Now put the value of ‛n = 5’ in equation(2), we get

=> Sn = 10 × 5 - 5

=> Sn = 50 - 5

=> <u>Sn</u><u> </u><u>=</u><u> </u><u>4</u><u>5</u><u>m</u>

\rm{\pink{\therefore}} The height of the journey is “ <u>4</u><u>5</u><u>m</u> ” .

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