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anonymous
 one year ago
A ballplayer catches a ball 3.18s after throwing it vertically upward. With what speed did he throw it, and what height did it reach? Can you show me the steps too!
anonymous
 one year ago
A ballplayer catches a ball 3.18s after throwing it vertically upward. With what speed did he throw it, and what height did it reach? Can you show me the steps too!

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do you have the answers? Just so we could match them?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't have the answers :(

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay so the total time is 3.18s divide the time by 2 and you'll get the time taken to reach the maximum hight and that is when v=0

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The inital speed when ball takes off is the same as when it reaches ballplayer's hands . but opposite in direction

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So t=1.59s v=0 u=? And g= 9.81m/s^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0By putting the formula v=ugt * note i used negative sign because we're going against the gravity. So 0=u(9.81×1.59} you'll get initial velocity

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now for maximum height use the formula s=ut1/2gt^2 you'll get s the maximum height.
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