43. We neglect air resistance for the duration of the motion (between “launching” and “landing”), soa=−g=−9.8m/s2(we take downward to be the−ydirection). We use the equations in Table 2-1 (with∆yreplacing ∆x) because this isa= constant motion.(a) At the highest point the velocity of the ball vanishes. Takingy0=0,wesetv=0inv2=v20−2gy,and solve for the initial velocity:v0=√2.Sincey= 50 m we Fndv0=31m/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 timet>0) we havey=v0t−12gt2=⇒t=2v0gwhich (using our result from part (a)) producest=6.4 s. It is possible to obtain this withoutusing part (a)’s result; one can Fnd the time just for the rise (from ground to highest point) from
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