Help on physics problem

[sebosfato]

I would need some help here

given g=10m/s (gravity acceleration)

How much time a body takes to drop from 5m hight?

I was trying

H=acceleration*timesquared divided by two

but i end with the square root of 1 witch is 1 so wrong

can anyone help me, what i'm doing wrong?

Is the formula wrong for 5m situation?


S=V0*t+a*t^2/2
???

thanks

[sebosfato]

I tried with a graphic and i found that in 0,5 second would make 3,33 meters

but how about 5 meters?

i'm totally confused

[quantumuppercut]

lol bro

maybe it's 1 second?

try potential kinetic energy method

mgh=1/2mv^2 ===> 2gh=v^2

1.41 * 50^1/2 = v

v= 1.4 * 7 = 9.8 m/s ?

Double check: Vf=Vo + at lol

try 1/3 + 1/3 +1/3 = 1 = .3333... x 3 = .9999... = 1?

[p75213]

How about this:
PE_initial=KE_final
mgh=1/2mv^2;
v=sqr(2gh)
average v=sqr(2gh)/2
h/t=sqr(2gh)/2 ========>v=x/t
t=2h/sqr(2gh)
t=2*5/sqr(2*9.81*5)
Ans=1.01 secs

[sebosfato]

Hello,

thanks guys,

I appreciate the help.

I found the problem cause i was thinking that the gravity acceleration is 10m/s and by consequence something dropping from 10m hight should hit the ground in exactly 1 second. But when i used the scalar calculation using 5meters i ended with the square root of 1 witch is 1, seems really stupid problem and probably really is...

I just tried again to re do the graphic now with fresh mind and found the problem.

The speed will reach 10m/s if the thing drops for 1 second ok but it will travel only 5meters because is the area of the graphic. The 10m/s will be only the speed it will reach in 1 second and not the distance it will travel.

check the graph i made

I was very tired and could not visualize this.

I thank you a lot for the help quantumuppercut and p75213.

I'm studying this cause i have a test to get a place on the university within 2 weeks.. hope to pass it.. i'm trying to start physics next year.

Thanks again

[Armagdn03]

change in distance = .5(a)(t^2)+Vit

a= acceleration (due to gravity)
t = time

or.... change in distance = one half acceleration times time squared plus the initial velocity times time.

Generally initial speed is zero so you can eliminate Vit.

so you would have 5m=.5(-9.81)(t^2)

so solve for t. should be around 1 second.

[p75213]

May I suggest:

Physics Help Forum