Calculating Time Taken for an Object Dropped from 40 m & Weighing 53kg

[Turkish]

Okay, I need some help working out the time taken for an object to fall, please Can you tell me how its done, with the working out, Thanks in advance.
This is the question:
An object is dropped from 40metres and weighs 53kg, Whats the time taken for it to hit ground.

 

[HallsofIvy]

Weight (or mass, which is really what you give) is irrelevant. Assuming no air resistance (which I have to since you didn't give any coefficient for resistance) acceleration is constant: -9.8 m/s².

The velocity after t seconds is -9.8t m/s and the distance fallen is the negative of the integral of that: 4.9t² m.

4.9t²= 40 gives t²= 40/4.9= 8.16 so t= √(8.16)= 2.8 seconds.

 

[Turkish]

Thanks for that, Really appreciate it, however I am a bit confused on how you managed to get 4.9:S

 

[ponjavic]

The integral of -9.8t would be -9.8t^2/2=-4.9t^2 then you were supposed to take the negative which equals 4.9t^2

 

[jamie_o]

I would use one of those 4 equations of motion. If you take
v = final velocity
u = initial velocity
a = acceleration
s = displacement
t = time taken

s = ut + 1/2at^2
sign convention i think its called means you can keep downwards to be the positive direction.
v = not needed
u = 0
a = 9.8
s = 40
t = ??

40 = 0*t + 1/2*9.8*t^2

40 = 4.9t^2

40/4.9 = t^2 so square root of 40/4.9 = t which is equal to 2.9seconds.

(That equation is derived from a = (v-u)/t and the area under the curve of a velocity/time graph)

Hope this helps.