Please help! A ball is thrown vertically upward. What is its acceleration after it has left the thrower's hand and is traveling upward? Also, at the instant it reaches the top of its flight and on its way down?

Question

anonymous · Answer

The answers on the back of the book states that it is 9.8 m/s^2 [down], and I'm aware that 9.8 m/s^2 is the acceleration due to the gravity.

amistre64 · Answer

acceleration is constant; so it doesnt matter where its at its always being affected at a constant rate

anonymous · Answer

ohhh, i see. And the constant rate is the gravity's pull?

amistre64 · Answer

acceleration is a change in speed; and its always chainging its speed at a constant rate .... yep

anonymous · Answer

That's not a very good way of explaining it...

Gravity is always acting on upon the ball, but by that reasoning it's accelerating when it's in the boy's hand as well. The acceleration of the ball is a result to the "net" or sum of the forces; since gravity is the only significant force, the only acceleration we consider is gravity's. The acceleration isn't actually constant, but locally it's close enough that we don't worry about it. You'll learn how gravity varies after you finish force and kinematics :)

amistre64 · Answer

Why would I concern myself with any definition of acceleration due to gravity while its in his hand, as the question asks "hat is its acceleration after it has left the thrower's hand".

With out writing a chapter about gravity and all its nuances, I figure the asker has the wherewithal to comprehend the answer given.  Just saying ....