Acceleration of a Ball: Understanding the Force of Gravity

  • Thread starter Thread starter simplekin93
  • Start date Start date
  • Tags Tags
    Acceleration Ball
AI Thread Summary
The acceleration of a ball thrown upwards is consistently -9.8 m/s² due to Earth's gravity, regardless of its upward motion. This value represents the acceleration acting on the ball, not the force, and remains constant throughout its flight. While the ball's velocity decreases as it rises, the acceleration remains directed downward. At its peak, the ball's velocity is zero, but it still experiences this downward acceleration. Understanding that acceleration can differ in direction from velocity is crucial in grasping these concepts.
simplekin93
Messages
5
Reaction score
0

Homework Statement


You throw a ball straight up with a speed of 14m/s. What is the acceleration of the ball 1.5s after you threw it?

Alright, the dilemma I am having is the answer is -9.8m/s^2.

It doesn't make sense to me how the acceleration can be that at 1.5s. The -9.8m/s^2 (gravity of earth) is the force pulling down on the ball not the acceleration after 1.5S as it is still moving up rather then down. What is the correct acceleration, or is -9.8m/s^2 correct?
 
Physics news on Phys.org
-9.8 is correct because the time doesn't matter. Gravity will have a constant acceleration on the object. 9.8 is the acceleration, NOT FORCE, because the unit for acceleration is m/s^2
 
-9.8ms-2 is correct. It looks to me like you have velocity and acceleration confused. Velocity is the speed and direction that the ball is moving, acceleration is the speed and direction in which the velocity is changing. This can be a tough concept to grasp (I found it a little difficult at least).

On the ball's way up, every second its velocity will be becoming 9.8ms-1 slower. When it's at its peak, its velocity will be zero, however it can still have an acceleration (9.8ms-2). Then on its way down, it will be speeding up by 9.8ms-1 every second.

From this we can see that the velocity of the ball has completely changed direction, whilst the acceleration hasn't changed at all!

I am aware that I'm not great at explaining things like this, but I hope it's helped a bit until someone else can come and explain the concept better!

Two important things to note are:
acceleration does not have to be in the same direction as velocity, if it's in the opposite direction, it merely means that it's slowing down.
acceleration due to gravity (at levels we're dealing with) does not change.
 
arlight thanks, that clears it up!
 
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 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
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