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

Question

A 1.00 kg 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, Ug(h), 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.

Please answer and explain! Thank you!!

Part A Draw a line on the graph representing the total energy Eof the bal. +add graph-add Points! Ureset?help d delete graph graph into E (J) 150 p 125 100 75 50 25 5.0 10.0 15.0 Submit Hints MyAnswers Give Up Review Part

Explanation / Answer

Total Energy = 1/2*mv^2 + m*g*H

Initially,
P.E = 0
K.E = 1/2 * 1.0 * 16^2
K.E = 128 J

This will be the total Energy of the system, Please draw a Horizontal line at E = 128 J mark.
That will be your answer.

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