Chapter 8: Problem 61
A ball of mass \(m=0.75 \mathrm{kg}\) is thrown straight upward with an initial speed of \(8.9 \mathrm{m} / \mathrm{s}\). (a) Plot the gravitational potential energy of the block from its launch height, \(y=0,\) to the height \(y=5.0 \mathrm{m} .\) Let \(U=0\) correspond to \(y=0\) (b) Determine the turning point (maximum height) of this mass.
Short Answer
Step by step solution
Define the potential energy equation
Calculate potential energy at different heights
Understand the implication of U=0
Determine maximum height using energy conservation
Solve for maximum height
Plot the potential energy against height
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Kinetic Energy
- \(m\) is the mass of the object
- \(v\) is the velocity of the object
Energy Conservation
Projectile Motion
- Ascent: The ball goes up, decelerating because gravity opposes its motion upward.
- Descent: After reaching the peak, the ball starts descending, reversing its motion direction and accelerating back down due to gravity.