an object is thrown straight up at 10 m/s, at the top of its rise what is the magnitude of the object's acceleration
The ONLY FORCE on this object is F = weight = -m g
so
about 9.81 m/s^2 downward (on earth)
When an object is thrown straight up, the only force acting on it is gravity. Gravity always acts in the downward direction, so at the top of its rise, the object's velocity is momentarily zero, and it starts to fall back downwards.
The magnitude of the object's acceleration at the top of its rise is equal to the acceleration due to gravity, which is approximately 9.8 m/s² on Earth. Therefore, the magnitude of the object's acceleration at the top of its rise is 9.8 m/s².
To determine the magnitude of the object's acceleration at the top of its rise, we need to understand the motion of the object.
When the object is thrown straight up, it experiences a constant acceleration due to gravity in the downward direction. This acceleration is commonly denoted as g and has a magnitude of approximately 9.8 m/s^2 on Earth.
At the top of its rise, the object momentarily stops before reversing its direction and falling back to the ground. During this brief moment, the object's velocity becomes zero.
Since the object's velocity changes from a positive value (upward) to zero velocity, the acceleration at the top of the rise is determined by the downward acceleration due to gravity, which is g.
Therefore, the magnitude of the object's acceleration at the top of its rise is approximately 9.8 m/s^2.