A juggler throws a ball with an initial horizontal velocity of +1.1 m/s and an initial vertical velocity of +5.7 m/s. What is its acceleration at the top of its flight path? Make sure to consider the sign when responding. Consider the upward direction as positive.
-g the whole time
What equation would I use for this?
To find the acceleration at the top of the flight path, we need to consider the forces acting on the ball and the motion it undergoes.
The only force acting on the ball in the vertical direction is the force of gravity. Near the surface of the Earth, the acceleration due to gravity is approximately 9.8 m/s² and acts downward. Therefore, the acceleration due to gravity for our calculation will be -9.8 m/s², considering the upward direction as positive.
At the top of the flight path, the ball reaches its maximum height and momentarily comes to rest before falling back down. When the ball is at the top, its vertical velocity is zero since it's not moving up or down. Therefore, the acceleration at the top of the flight path is solely due to gravity and can be considered as -9.8 m/s² in the upward direction.
So, the acceleration at the top of the flight path is -9.8 m/s² upwards.