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Answer to Question #14913 in Mechanics | Relativity for mohamed alarayyed
Question #14913
A baseball is thrown horizontally from a height of 122.5 m above the ground with a speed of 26.5 m/s. Where is the ball after 5.00 s has elapsed?
horizontal
m from the launch position
vertical
m from the ground
horizontal
m from the launch position
vertical
m from the ground
Expert's answer
horizontal m from the launch position
L=((26.5)^2)*122.5/10=8602.56
vertical m from the ground
L=((26.5)^2)/20=25.11
L=((26.5)^2)*122.5/10=8602.56
vertical m from the ground
L=((26.5)^2)/20=25.11
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