A stone is thrown vertically upwards, reaches a highest point, and returns to the ground. When the stone is at the top of its path, its acceleration
Now Alex, enough already, draw picture with all forces.
There is one, the weight
F = - m g
so
F = m a = - m g
so
a = -g
THE WHOLE TIME, NOT JUST AT TOP
Now no more questions until you try yourself !!!!
When the stone is at the top of its path, its acceleration is equal to the acceleration due to gravity, which is approximately 9.8 m/s². This acceleration is directed downwards, opposing the motion of the stone. At the highest point, the stone momentarily comes to rest before starting to fall back to the ground.
To determine the acceleration of the stone when it is at the top of its path, we can use the equation for acceleration in free fall:
acceleration = (final velocity - initial velocity) / time
In this case, when the stone is at the top of its path, its final velocity is 0 m/s because it momentarily stops before falling back down. We can assume that the initial velocity is also 0 m/s when the stone is thrown vertically upwards.
Now, let's consider the concept of acceleration due to gravity. When the stone is thrown upwards, gravity acts against its motion, slowing it down. As it reaches the top and starts falling downward, gravity accelerates the stone towards the ground, increasing its speed.
The acceleration due to gravity on Earth is approximately 9.8 m/s^2, directed downwards. So, at the top of its path, the stone experiences an acceleration of 9.8 m/s^2 in the downward direction.
Therefore, when the stone is at the top of its path, its acceleration is approximately 9.8 m/s^2 downwards due to the acceleration of gravity.