The Physics of Throwing a Ball: Velocity & Acceleration

AI Thread Summary
When a ball is thrown upwards, its velocity upon descent is not the same due to air resistance, which slows it down. If air resistance is neglected, the speed at the same vertical height during ascent and descent is equal, but the direction of the velocity changes. The acceleration due to gravity remains constant at 9.8 m/s² throughout the motion. Thus, while the speed can be the same at a specific height, the overall velocity differs because of direction. Understanding these principles clarifies the physics of projectile motion.
Maxwell
Messages
511
Reaction score
0
Just a quick question -

If I throw a ball up in the air, will the velocity be the same coming down as it was going up? Also, will the acceleration change?

Why does this happen?

Thanks.
 
Physics news on Phys.org
Maxwell said:
Just a quick question -

If I throw a ball up in the air, will the velocity be the same coming down as it was going up?
No, it won't. You said "up in the air", so there will be air resistance slowing it down and it will come down more slowly than it went up. To get the idea, throw a badminton birdie or a feather up as hard as you can. For baseballs, and small thrown heights, however, the difference may not be noticeable.
 
What if you neglect air resistance?
 
If you neglect air resistance the ball will be at the same speed as it comes past the place where it was released upwards just going in the other direction and the acceleration is still g, 9.8 m/s^2
 
The acceleration doesn't change. I'm assuming that you're still on the surface of the Earth of course.
 
The acceleration due to gravity never changes and is always 9.8 m/s2. Neglecting air resistance, we can therefore reason that the speed of the ball at a certain vertical height when thrown up is the same as its speed at the same vertical height when it falls to Earth after being thrown. The velocity, however, changes because the ball is no longer moving in the same direction.
 
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