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
This is the end of the preview. Sign up
access the rest of the document.
This note was uploaded on 09/03/2009 for the course PHYS 1200 taught by Professor Stoler during the Spring '06 term at Rensselaer Polytechnic Institute.