Given: h = 600 m, the height of descent t = 5 min = 5*60 = 300 s, the time of descent.
Let a = the acceleration of descent., m/s². Let u = initial velocity of descent, m/s. Let t = time of descent, s. The final velocity is v = 0 m/s because the helicopter comes to rest on the ground. Note that u, v, and a are measured as positive upward.
Then u + at = v (u m/s) + (a m/s²)*(t s) = 0 u = - at u = - 300a (1)
Also, u*t + (1/2)at² = -h (um/s)*(t s) + (1/2)(a m/s²)*(t s)² = 600 ut + (1/2)at² = 600 (2) From (1), obtain -300a +(1/2)(a)(90000) = -600 44700a = -600 a = - 1.3423 x 10⁻² m/s²
From (1), obtain u = - 300*(-1.3423 x 10⁻²) = 4.03 m/s
Answer: The acceleration is 0.0134 m/s² downward. The initial velocity is 4.0 m/s upward.
