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A 3.1 kg ball is thrown straight upward with a speed of 18.2 m/s. Use conservation of energy to calculate the maximum height the ball can re
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Answers ( )
Answer:
h = 16.9 m
Explanation:
When a ball is thrown upward, its velocity gradually decreases, until it stops for a moment, when it reaches the maximum height, while its height increases. Thus, the law conservation of energy states in this case, that:
Kinetic Energy Lost by Ball = Potential Energy Gained by Ball
(0.5)m(Vf² – Vi²) = mgh
h = (0.5)(Vf² – Vi²)/g
where,
Vf = Final Speed of Ball = 0 m/s (Since, ball stops for a moment at highest point)
Vi = Initial Speed of Ball = 18.2 m/s
g = acceleration due to gravity = – 9.8 m/s² ( negative for upward motion)
h = maximum height the ball can reach = ?
Therefore, using values in the equation, we get:
h = (0.5)[(0 m/s)² – (18.2 m/s)²]/(-9.8 m/s²)
h = 16.9 m