Answer to Question #14913 in Mechanics | Relativity for mohamed alarayyed

Question

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

Answer 1

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