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a ball is thrown straight downward from the top of a tall building. the initial speed of the ball is 10m/s. it strikes the ground with a speed of 60m/s. how tall is the building?
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Well the height of the building is the distance the ball covers. You have the initial velocity, the final velocity, the acceleration (which is the gravitational acceleration, 9.8 m/s squared). So you use this equation:\[v _{f}^{2}=v _{i}^{2}+2ax \]Then you solve for x (the distance traveled).\[x=\frac{ v _{f}^{2}-v _{i}^{2} }{ 2a }=\frac{ (60\frac{ m }{ s })^{2} -(10\frac{ m }{ s })^{2}}{ 2 \times 9.8\frac{ m }{ s ^{2} } }=130m\]The answer is 130m aprox.
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