Question

A small bag of sand is released from an ascending hot‑air balloon whose constant, upward velocity is ?0=2.85 m/s . Knowing that at the time of the release the balloon was 60.8 m above the ground, determine the time it takes for the bag to reach the ground from the moment of its release. Use ?=9.81 m/s2.

Answer 1

Given: V=2.85mys and, h= 60.8m let the height reached by the bag above height his k. h'= V² = (2,85 m/s) 2 = 0.414m. 29 269.81ms”) and time taken to reach maximum height, t= Vo= (20 85 Mys) = 0.29 sec g (9.81 m/s²) Here, Maximum height of bag above ground is; H=hth. ..H= 60.8m + 0.414 m = 61.214.m so, time taken by bag to reach to the ground from this height; t'= 24- 2161.214) = 3.53 sec. g J (9.81) Therefore; Total time required is: Tot + ť ::T= (0.29s)+(3.53 s) ET=3.82 sec