Question

00:31:17 A person standing at the edge of a cliff throws a rock upward from a height of 400 ft above ground level with an initial velocity of 62 ft/s. What is the velocity of impact when the rock reaches the bottom of the cliff? Round any intermediate calculations, if needed, to no less than six decimal places, and round your final answer to two decimal places. (Hint: Use h(1) = 1612 + 621 + 400 as the position function, where his in feet, e in seconds. Ignore air resistance.) Answer 2 Points < Keypad Keyboard Shortcuts > Prev Next ft/s

Answer 1

8 I cliff >> ta v2 = Solution :- given, initial velocity, u = 62 of 4/5 acceleration due to gravity? g=32.17.405 ft/s2. at the point when dock reach upper most point final velocity: v=u- gxt at upper most point :v=o 0 = U-97 > git=u»trų 900 feet 62 -1.9 27018 gooand 39.17405 this is the time in which rock reaches point B. and u 2 - 2 aos is distance b/w A and B. > or (62) 2*22./7405x S (62)2 59.73758 2X 32.17405 " height b c = 400 +59.73758 = 459.73758. for be; use equation of motion : v2 = u2t2.a.s for Be : u= velocity at V = velocity at point c. >> V2 = 0)2+28 32.17405 X 459.7 2758 171.998 ft/second Am so, velocity of impact is 171.998 ft/second when the rock reaches the bottom of Cliff. Ans. at point B = 0 V