Projectile Motion: Time and Impact Velocity Comparison - Homework Question

AI Thread Summary
When a ball is tossed upward and then allowed to fall, the time to reach the ground and the impact velocity are not equal due to the different starting heights. However, in ideal physics scenarios without air resistance, both trajectories yield the same impact velocity when calculated. The discussion emphasizes that while air resistance is often ignored in basic projectile motion problems, it significantly affects real-world outcomes. If air resistance is considered, the times and velocities would differ for the two scenarios. Understanding these principles is crucial for accurately analyzing projectile motion.
phizics09
Messages
38
Reaction score
1

Homework Statement



When you toss a ball up and let it drop to the ground, compared to when you just let it fall, will the time it takes to reach the ground and impact velocity be equal?

Homework Equations





The Attempt at a Solution


For the time, they shouldn't be equal, since the height from which the ball starts falling is different right? But how come when you solve projectile motion problems with an original velocity in the vertical direction, you don't need to calculate the separate two paths--one up and one down--separately, but you still get the same answer?
 
Physics news on Phys.org
An assumption (physics is full of 'good' assumptions) is that air resistance can be ignored.
This is good in the first instance but it is obviously untrue and must be taken into account in a full analysis. But, you have to start some where
 
What do you mean?
 
Think about the two trajectories. In the case where the ball is thrown straight up, it will eventually reach a highest point at which its height from the ground is h. What is the velocity of the ball at that point? How is the similar or different to the velocity of the ball if you were to simply drop it from that same height?
 
If there is no air resistance then there will be no difference at all
 
technician said:
If there is no air resistance then there will be no difference at all

Now if you throw in air resistance, and assume the balls are identical, think about what would happen to the times then.
 
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