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If a ball leaves the ground with a velocity of 4.67 m/s,
how high does the ball travel?

User Ujjwal
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1 Answer

5 votes

Answer:


Vf^2=Vo^2+2aS\\(0m/s)^2=(4.67m/s)^2+(2*-10m/s^2)S\\-(4.67)^2 m^2/s^2=-20m/s^2*S\\S=(21.8089/20) m\\S=1.090445 m\\

User Bravokiloecho
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