Air Drag & No Drag on a Ball Projected Vertically

AI Thread Summary
The discussion focuses on the differences in velocity-time (v-t) graphs for a ball projected vertically with and without air drag. In the case without air drag, the velocity decreases linearly from the initial velocity V to 0, represented by a straight line with a slope of -g. When air drag is considered, the motion is more complex; the velocity decreases exponentially and approaches terminal velocity rather than following a linear path. The participants clarify that the v-t graph remains a straight line only under constant acceleration, which is not the case with air drag. Ultimately, the discussion concludes with the realization that the presence of air drag alters the motion significantly, leading to a non-linear relationship.
AdityaDev
Messages
527
Reaction score
33
< Moderator Note -- thread moved from General Physics to Homework Help forum >[/color]

If a ball is projected vertically upwards say with a velocity V.

Case 1: Without air drag
Case 2: With air drag

What will be the difference in v-t graphs and the motion of the ball
Air drag D=bv (b is a constant,v is velocity)

I know that in case 1, the velocity decreases as the ball goes up from V to 0.
##v=V-gt ##
##(y=-mx+c)##
Hence it will be a straight line with slope -g.
With air drag will the graph still be a straight line??
 
Last edited by a moderator:
Physics news on Phys.org
AdityaDev said:
< Moderator Note -- thread moved from General Physics to Homework Help forum >[/color]

If a ball is projected vertically upwards say with a velocity V.

Case 1: Without air drag
Case 2: With air drag

What will be the difference in v-t graphs and the motion of the ball
Air drag D=bv (b is a constant,v is velocity)

I know that in case 1, the velocity decreases as the ball goes up from V to 0.
##v=V-gt ##
##(y=-mx+c)##
Hence it will be a straight line with slope -g.
With air drag will the graph still be a straight line??
What does the slope of the v-t graph represent? And when is this graph a straight line?
 
PeroK said:
What does the slope of the v-t graph represent? And when is this graph a straight line?
it is a straight line.. for consrant acceleration, vt graph is a straight line which is inclined... v=-gt+V is a straight line. pls don't say its parabolic.
 
And is there constant acceleration with air drag?
 
PeroK said:
And is there constant acceleration with air drag?
i got the answer...i was talking about no air drag case...with air drag, after serious integration, i found v(t) and it is a function with exponent and is in terms of terminal velocity...you can close the thread.
 
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