Question

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.​...

Answer 1

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

• 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 = 45m -----(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 = 5

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

=> Sn = 10 × 5 - 5

=> Sn = 50 - 5

=> Sn = 45m

$\rm{\pink{\therefore}}$ The height of the journey is “ 45m ” .