Distance fallen after 3 seconds? 5 seconds?

Question

If a person jumps from a building and maintains a verticle position with arms at the side how far would they have fallen after 3 seconds? 5? 8? Conversely, how fast would you be going when you hit the ground after jumping from a 5 story building? 10 story? and how long would those last two trips take?

Answer 1

It depends on the initial vertical speed. However, if you assume that it is zero, then you can calculate the falling distance and speed by the two fomulae:
d = g*t^2/2 and
v = g*t
where g is the acceleration of gravity and is (approximately) equal to 10 m/s or 32.8 ft/s, t is time, d is distance and v is velocity.

Then: 3 seconds: d = 45 m = 147.6 ft
5 seconds: d= 125 m = 410 ft
8 seconds: d= 320m = 1050 ft
(Imagine how incredibly fast it is!)

Now to the second question: suppose a story is 3 m high, then a 5 story building is 15m = 10 * T^2/2, where T is the falling time. Then T = squareroot(15/10*2) = 1.732

Then you would be falling as fast as: v = g*T ~= 17.3 m/s = 56 ft/s.
You can do the same for the rest and it is ~ 24.5 m/s = 80.4 ft/s

Sorry for my long answer

Answer 2

Distance Fallen

Answer 3

Rodney_Lee has given an clever answer which I examine when I had responded. He needs to spectacular typo errors in his final equation from a hundred*b to a hundred*t. whether, that's a mathematical answer which represents a hypothetical difficulty. right here, easily the object isn't overlaying 6 ft. in 0 2d, yet time is counted after it has coated 6 2d. exciting factor is that the final greater desirable suggestion gets confirmed, it shows that the instructor became additionally thinking on a similar line. in certainty, countless such equations might properly be formed utilising all of the 4 instruments of advice. yet while 3 instruments of advice is used and the final suggestion is used to envision, then no longer all of those countless suggestions will greater healthful the final suggestion. If Rodney_Lee is stable in his answer, your instructor must be coaching you a financial ruin on sequence and sequences in math. If i'm spectacular, then your instructor ought to have been coaching you a financial ruin of viscosity in physics. My answer is as below. This must be a parachute which became opened after it had fallen 6 ft. In calculating time, digits after decimal factor are ignored. So it takes 0 sec. Then it covers lesser and lesser diatance according to 2d, because of the fact because of drag tension performing against it increasing with its increasing velocity. So the internet downward tension and acceleration keeps lowering ensuing in it overlaying lesser and lesser diatance with time. ultimately, that's going to attain a terminal velocity which would be consistent.

Answer 4

Theoretically an object without much air resistance increases its speed by 32 feet per second every second. So an object will reach 100 feet per second just before it hits the ground if it is sent at 400 feet.

Answer 5

To find the speed multiply the time squared by 4.9. 44.1metres, 122.5 metres and 313.6 metres
What is the height of this imaginary 5 or 10 storey building?

Answer 6

distance in feet s=.5 g t^2

where t is time in sec
g=32 is the gravitational constant

depends on how much height a "story" is For height h

h==.5 g t^2

solve for t
t=sqrt(2h/g)