A small mailbag is released from a helicopter that is descending steadily at 2.9

Question

A small mailbag is released from a helicopter that is descending steadily at 2.98 m/s. (A) after 5.00s, what is the speed of the mailbag? (B) how far is it below the helicopter? (C) what are your answers to part (a) and (b) if the helicopter is rising steadily at 2.98 m/s? v = d = A small mailbag is released from a helicopter that is descending steadily at 2.98 m/s. (A) after 5.00s, what is the speed of the mailbag? (B) how far is it below the helicopter? (C) what are your answers to part (a) and (b) if the helicopter is rising steadily at 2.98 m/s? v = d = (A) after 5.00s, what is the speed of the mailbag? (B) how far is it below the helicopter? (C) what are your answers to part (a) and (b) if the helicopter is rising steadily at 2.98 m/s? v = d =

Explanation / Answer

v0 = 2.97 m/s (downward and taking downward direction as +ve direction)

a = 9.8 m/s^2

(A) v = v0 + a t

v = 2.97 + (9.8 x 5) = 51.97 m/s

(B) d = v0 t + a t^2 / 2

d = (2.98 x 5) + (9.8 x 5^2 / 2)

d = 137.4 m

(C) then v0 = - 2.98 m/s

v = - 2.98 + (9.8 x 5) = 46.02 m/s

  
d = (-2.98 x 5) + (9.8 x 5^2 / 2)

d = 107.6 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.