A small mailbag is released from a helicopter that is descending steadily at 2.0

Question

A small mailbag is released from a helicopter that is descending steadily at 2.00 m/s. (a) After 5.00 s, what is the speed of the mailbag? v = 51 m/s (b) How far is it below the helicopter? d = 132.5 Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation carry accuracy to minimize roundoff error. m (c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at 2.00 m/s? v = 47 m/s d = 112.5 Your thin 10%ofthe correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out accuracy to minimize roundoff error. m

Explanation / Answer

Given,

a) Initial speed of the mail bag, u = 2 m/s

Final velocity after 5 sec, v = u + at = 2 + 9.8 x 5 = 51 m/s

Distance traveled by mail bag in 5 sec, Sm = ut + 0.5at2 = (2 x 5) + (0.5 x 9.8 x 52) = 132.5 m

Distance traveled by mail helicopter in 5 sec, Sa = 2 x 5 = 10 m

So, difference in the distance = 132.5 - 10 = 122.5 m

So, the mail bag is 122.5 m below the helocopter

b) Helicopter is ascending

Initial speed of the mail bag, u = -2 m/s

After 5 sec its speed, v = u+at = -2 + 9.8 x 5 = 47 m/s

Distance traveled by mail bag in 5 sec, Sm = ut+ 0.5at2 = -2 x 5 + 0.5 x 9.8 x 52 = 112.5 m

Distance traveled by mail helicopter in 5 sec, Sa = -2 x 5 = -10 m

So difference in the distance = 112.5 - (-10 ) = 122.5 m

The mail bag is 122.5 m below the helocopter

Please rate my answer if you find it helpful, good luck...

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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