Gravitational potential energy:

$\overline{)\begin{array}{rcl}{\mathbf{U}}& {\mathbf{=}}& \mathbf{m}\mathbf{g}\mathbf{h}\end{array}}$

Maximum height:

$\overline{){{\mathit{H}}}_{\mathit{m}\mathit{a}\mathit{x}}{\mathbf{=}}\frac{{{\mathit{v}}_{\mathbf{0}}}^{\mathbf{2}}}{\mathbf{2}\mathit{g}}}$

h = v_{0}^{2}/2g = 16.0^{2}/(2 × 10) = **12.8 m**

U_{max} = mgh = 1.00 × 10 × 12.8 = 128 J

A 1.00kg ball is thrown directly upward with an initial speed of 16.0 m/s.

A graph of the ball's gravitational potential energy vs. height, for an arbitrary initial velocity, is given in Part A. The zero point of gravitational potential energy is located at the height at which the ball leaves the thrower's hand.

For this problem, take g = 10 m/s^{2} as the acceleration due to gravity.

Draw a line on the graph representing the total energy of the ball.

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Equations for Energy Conservation concept. If you need more Equations for Energy Conservation practice, you can also practice Equations for Energy Conservation practice problems.

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Based on our data, we think this problem is relevant for Professor Weiler's class at BU.