Why Is Displacement Only 4.9m After 1 Second in Free Fall?

AI Thread Summary
The discussion centers on the confusion regarding the displacement of a free-falling object after one second. The correct formula for displacement is s = 1/2 * a * t^2, where 'a' is acceleration and 't' is time. Using this formula, the displacement after one second is calculated as 4.9 meters, not 9.8 meters, because it accounts for the total distance fallen from rest. To find displacement over specific intervals, one must calculate the difference in total displacements at different times. Understanding this distinction clarifies why the displacement is only 4.9 meters after one second.
mirza21
Messages
3
Reaction score
0
Hi friends am new to Physics. Am facing a problem while caculating displacment.

Acceleration of free falling object is 9.8m/s/s. Am using formula 1/2*b*h to get the displacement.According to me if object acceleration is 9.8 m/s/s then after 1 second it's displacement should be 9.8 meters and after 2 second it should be positioned at 29.4 m.
If i use the formula to calculate displacement then it is positioned at 4.9 m at end of 1st second and a the end of 2nd second it 19.6 m. Please explain why object is positon at 4.9 m by using this formula.


At t = 1 s
d = (0.5) * (9.8 m/s2) * (1 s)2 = 4.9 m

At t = 2 s

d = (0.5) * (9.8 m/s2) * (2 s)2 = 19.6 m

At t = 5 s

d = (0.5) * (10 m/s2) * (3 s)2 = 44.1 m

For refrence see http://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L5d.html
 
Last edited by a moderator:
Physics news on Phys.org
mirza21 said:
Hi friends am new to Physics. Am facing a problem while caculating displacment.

Acceleration of free falling object is 9.8m/s/s. Am using formula 1/2*b*h to get the displacement.According to me if object acceleration is 9.8 m/s/s then after 1 second it's displacement should be 9.8 meters and after 2 second it should be positioned at 29.4 m.
If i use the formula to calculate displacement then it is positioned at 4.9 m at end of 1st second and a the end of 2nd second it 19.6 m. Please explain why object is positon at 4.9 m by using this formula.


At t = 1 s
d = (0.5) * (9.8 m/s2) * (1 s)2 = 4.9 m

At t = 2 s

d = (0.5) * (9.8 m/s2) * (2 s)2 = 19.6 m

At t = 5 s

d = (0.5) * (10 m/s2) * (3 s)2 = 44.1 m

For refrence see http://www.glenbrook.k12.il.us/gbssci/phys/Class/1DKin/U1L5d.html

The statement in red is totally wrong.It also depends on the initial velocity of the object from which it was thrown downwards.
NOW for the hint part:

It seems you are confused with the formula part.
The formula s=1/2 at^2 [u=0] does not give the displacement in that particular time interval,in fact It gives you the total displacement from the initial time interval.
To get the displacement in a particular time interval you must calculate the displacements using the formula and substituting the correct values and then calculate the difference.
 
Last edited by a moderator:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top