(c) Using 2202vvgy=−(with different values for yand vthan before), we solve for the value of ycorresponding to maximum height (where v= 0). 2202(3.7 m/s)0.698 m.(9.8m/s)vyg===Thus, the armadillo goes 0.698 – 0.544 = 0.154 m higher. 48. We neglect air resistance, which justifies setting a= –g= –9.8 m/s2(taking downas the –ydirection) for the duration of the motion. We are allowed to use Table 2-1 (with Δyreplacing
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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.