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!
Do you have the answers? Just so we could match them?
@mewi2671
I don't have the answers :(
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
The inital speed when ball takes off is the same as when it reaches ballplayer's hands . but opposite in direction
So t=1.59s v=0 u=? And g= -9.81m/s^2
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
Now for maximum height use the formula s=ut-1/2gt^2 you'll get s the maximum height.
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