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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.​...

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BrainlyHandsome

[tex]\rm{\gray{\underline{\underline{\blue{GIVEN:-}}}}}[/tex]

  • 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 .

[tex]\rm{\gray{\underline{\underline{\blue{TO\:FIND:-}}}}}[/tex]

  • The height of the journey .

[tex]\rm{\gray{\underline{\underline{\blue{SOLUTION:-}}}}}[/tex]

We have know that,

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

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 = 45m -----(1)

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

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

=> 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 = 5

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

=> Sn = 10 × 5 - 5

=> Sn = 50 - 5

=> Sn = 45m

[tex]\rm{\pink{\therefore}}[/tex] The height of the journey is “ 45m ” .

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