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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!

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  1. anonymous
    • one year ago
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    Do you have the answers? Just so we could match them?

  2. anonymous
    • one year ago
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    @mewi2671

  3. anonymous
    • one year ago
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    I don't have the answers :(

  4. anonymous
    • one year ago
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    Okay 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

  5. anonymous
    • one year ago
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    The inital speed when ball takes off is the same as when it reaches ballplayer's hands . but opposite in direction

  6. anonymous
    • one year ago
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    So t=1.59s v=0 u=? And g= -9.81m/s^2

  7. anonymous
    • one year ago
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    By putting the formula v=u-gt * note i used negative sign because we're going against the gravity. So 0=u-(9.81×1.59} you'll get initial velocity

  8. anonymous
    • one year ago
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    Now for maximum height use the formula s=ut-1/2gt^2 you'll get s the maximum height.

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