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Physics 15 Online
OpenStudy (anonymous):

A ball is thrown horizontally at a speed of 24 meters per second from the top of the building. If the ball hits the ground 4.0 seconds later approximately how high is the building?

OpenStudy (anonymous):

Given that the problem said that the ball was thrown from the building horizontally means Vox=24 m/s and Voy=0. Through solving the path which the ball traveled by components we can solve for the height (I'll call it Yo): Yo is my initial height where the ball is coming from Y is our final height which is the ground so I'll call it 0. Y=Yo+Voy t- (1/2) gt^2 with that said Voy t = 0 since Voy=0, with Y=0, 0=Yo-(1/2)gt^2 Yo= (1/2)gt^2 Yo= (1/2)(9.8 m/s^2)[(4s)^2] Yo= 78.4 m

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