Work-energy theorem: A ball is thrown upward at an angle with a speed and direction such that it reaches a maximum height of 16.0 m above the point it was released, with no appreciable air resistance. At its maximum height it has a speed of 18.0 m/s. With what speed was the ball released?

Question

Quizplus · Accepted Answer

The work-energy theorem states that the change in kinetic energy of an object is equal to the work done on it. When the ball is at its maximum height, it has both potential energy due to its height and kinetic energy due to its speed. The initial kinetic energy when the ball was released can be found by adding the kinetic energy at the maximum height to the potential energy at the maximum height. Using the equations for kinetic energy ($KE = \frac{1}{2}mv^2$) and potential energy ($PE = mgh$), where $m$ is mass, $v$ is velocity, $g$ is the acceleration due to gravity (9.8 m/s$^2$), and $h$ is height, we can solve for the initial velocity.Given:- Maximum height ($h$) = 16.0 m- Speed at maximum height = 18.0 m/s- $g$ = 9.8 m/s$^2$At the maximum height, the potential energy is $PE = mgh = m \times 9.8 \times 16$, and the kinetic energy is $KE = \frac{1}{2}m \times (18.0)^2$.The total mechanical energy at the top is the sum of these, which must equal the initial kinetic energy since there's no air resistance and energy is conserved.Solving for the initial velocity involves setting the sum of the potential and kinetic energy at the top equal to the kinetic energy at the bottom (initial point) and solving for $v$. This calculation shows that the initial speed is approximately 25.3 m/s, which means the ball was released at this speed to reach the given conditions.

Work-Energy Theorem: a Ball Is Thrown Upward at an Angle