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

We have know that,

Where,

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

=> 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,

=> Sn = 0 + 10/2 (2 × n - 1)

=> Sn = 5 (2n - 1)

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

Therefore,

=> 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

The height of the journey is “ 45m ” .

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever

the current is 0.6

voltage ÷ resistance = current

so 12 ÷ 20 = 0.6