63. We choose downas the +ydirection and place the coordinate origin at the top of the building (which has height H). During its fall, the ball passes (with velocity v1) the top of the window (which is at y1) at time t1, and passes the bottom (which is at y2) at time t2.We are told y2– y1= 1.20 m and t2– t1= 0.125 s. Using Eq. 2-15 we have yyvttg tt2112121212−=−+−bgbgwhich immediately yields ()()221211.20 m9.8 m/s0.125 s8.99 m/s.0.125 sv−==From this, Eq. 2-16 (with v0= 0) reveals the value of y1:221112(8.99 m/s)24.12 m.
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