A ball is thrown straight up in the air
A) What is its velocity at the top of its path?
B) What is acceleration at the top of its path
A) v = 0,
B) a = g
To answer these questions, we need to have some understanding of the motion of the ball when it is thrown straight up in the air.
When the ball is thrown straight up, it experiences a constant acceleration due to gravity acting in the opposite direction to its motion. This acceleration is approximately 9.8 meters per second squared (m/s²), directed downward.
A) To determine the velocity of the ball at the top of its path, we can use the concept of conservation of energy. At the top of its path, the ball momentarily comes to a stop before reversing its direction and falling back down. At this point, all of its initial kinetic energy has been converted into potential energy. Therefore, we can say that the ball's velocity at the top of its path is 0 m/s.
B) The acceleration of the ball at the top of its path is also 9.8 m/s², directed downward, which is the same as the acceleration due to gravity. The acceleration remains constant throughout the ball's flight, regardless of its position.
In summary:
A) The velocity of the ball at the top of its path is 0 m/s.
B) The acceleration of the ball at the top of its path is -9.8 m/s² (directed downward).