47. We neglect air resistance for the duration of the motion (between “launching” and “landing”), so a= –g= –9.8 m/s2(we take downward to be the –ydirection). We use the equations in Table 2-1 (with Δyreplacing Δx) because this is a= constant motion. (a) At the highest point the velocity of the ball vanishes. Taking y0= 0, we set v= 0 in vvgy2022=−, and solve for the initial velocity: vgy02=.Since y= 50 m we find v0= 31 m/s. (b) It will be in the air from the time it leaves the ground until the time it returns to the ground (y= 0). Applying Eq. 2-15 to the entire motion (the rise and the fall, of total time
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