If a man weighs 90.75 kilograms, and he falls 7.75 meters, how long would it take him to fall in seconds?

Question

How hard would he hit the ground???

How hard would he hit the ground?

I know absolutely no physics and very little math, so please explain in terms a writer could understand if possible.

Answer 1

1.26 seconds (hint: his weight is irrelevant)

d = (initial velocity)(time) + (0.5)(acceleration)(time)squared
(7.75 m) = (0)t + (0.5)(9.8)t2
t = 1.26 s

Answer 2

Essentially, if you ignore air or wind resistance, the longer (farther) you fall the faster you fall, and the faster you fall the farther you fall in a given time. Without going through the derivations, the equations that apply here are
v = at
and
s = (1/2)at^2
a in this case is the acceleration due to gravity (usually designated g) which is 9.80665 meters per second per second.
Solving the 2nd equation for t (since we know s and a),
t^2 = 2s/a
t = √(2s/a)
t = √(2*7.75/9.80665)
t ≈ 1.257630 seconds ≈ 1.26 seconds
with t known, we can solve for v:
v = at = 9.80665*1.257630
v ≈ 12.32895 m/s ≈ 12.3 m/s or 44.4 kph
If there is nothing to cushion the impact, the odds of surviving are pretty low.

Answer 3

The weight doesn't matter.
The acceleration rate of gravity is 9.8m/second.
The first seconds you will fall 9.8m, the seconds second you will fall 19.6m, etc.........
Divide your known, 7.75 by 9.8 and this will give you the time in seconds.
You're falling less than the acceleration rate, so it will be less than a second.
I'll let you do the math.

Answer 4

There is not enough information in your question,the mans altitude must be known, the rate of acceleration depends on distance from the Earth's surface.

Answer 5

Who makes questions this sick?

First of all, mass does not figure into the equation. Second, the man is most likely dead.

Juck!

Answer 6

His weight doesn't matter. Use

y = 1/2gt^2 where g = 9.8 m/s^2

So t = sqrt(2y/g) = sqrt(2*7.75 m/9.8 m/s^2) = 1.26 seconds