Question

A cargo helicopter, descending steadily at a speed of 3.3 m/s, releases a small package. Let upward be the positive direction for this problem. (a) If the package is 31 m above the ground when it is dropped, how long does it take for the package to reach the ground? s (b) What is its velocity just before it lands? (Indicate the direction with the sign of your answer.) cm/s

Answer 1

Since final velocity is in downwards direction hence sign of Velocity must be negative.

TU= 3.3 m/s S = 31 m. Lu=? Given, u = 3.3 m/s S = 3im take a = 9.8 m/82 Using, 8 = ut + dat? > 31 = 3.3+ + 5*9.8x+ → 62 = 6.67 + 9.8+? 7 9.8+*+6.67-62 = 0 t = -6.6+ 6.6)24x9.8x6-62) 2x9.8 7 t = -6.6+ 49.73 19.6 + t = 2.2 secs. Ane Using v= utat → u = 3.3 + 9.8 * 2:2 le = 24.86 m/s 1 IL (6) Jane