Question about the acceleration of a ball thrown upwards?

g = 9.8 m/s²

d. directed downward. think of. what's inflicting it to strengthen up downward? Are any forces appearing upward on it? Can or not it is a threat for there to be a internet stress upward on it, after that's released from the hand and till now it impacts the floor? solutions: Gravity No No the internet stress could desire to be downward, and so could desire to the acceleration. The ball slows down because it ascends, and quickens because it descends. because of the fact its acceleration is often downward.

Hello Martin, the acceleration due to gravity ie g which equals to 9.8 m/s^2 is always on the body but acts always downward because it is directed towards the centre of the earth.

But when the body is moving up then acceleration is said to be deceleration and so the speed of the body goes on reducing and finally at one stage it comes to zero. Now the body would be at the maximum height.

As the body starts falling down then g would be the positive acceleration and so the speed goes on increasing as it comes down.

In a nut shell,

when the body is ascending the acceleration is -g and

when it is descending the acceleration is +g

The moment you release the ball, it's acceleration is g (downward.) You say "What?, that's impossible, the ball is still going up!" Yes, the ball is going up but the change in it's velocity (also known as acceleration) is g = -9.8m/s^2. Ignoring air resistance:

V(t) = Vi - g*t => a = dV/dt = -g

If you are asking about instantaneous acceleration then it is = -g = -9.81 m/s2

But if you are asking about average acceleration then

acc= (v-u)/t

since it lands back on the ground its final velocity v = 0

acc= -u/t

during upper journey acc= -g=v'-u/t'

where v' = 0 since its the topmost point of the journey and t'=t/2 the time for upper journey only

-g=-2u/t

Therefore -u/t=-g/2=-4.9 m/s2

Its average acceleration is 4.9 m/s2 downwards.

0 or -1

9.81 m/s^2 Down