anonymous
  • anonymous
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!
Physics
schrodinger
  • schrodinger
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anonymous
  • anonymous
Do you have the answers? Just so we could match them?
anonymous
  • anonymous
anonymous
  • anonymous
I don't have the answers :(

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anonymous
  • anonymous
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
anonymous
  • anonymous
The inital speed when ball takes off is the same as when it reaches ballplayer's hands . but opposite in direction
anonymous
  • anonymous
So t=1.59s v=0 u=? And g= -9.81m/s^2
anonymous
  • anonymous
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
anonymous
  • anonymous
Now for maximum height use the formula s=ut-1/2gt^2 you'll get s the maximum height.

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