physics homework question

Question

A small mailbag is released from a helicopter that is descending steadily at 1.10 m/s.

(a) After 5.00 s, what is the speed of the mailbag?

v = ___m/s

(b) How far is it below the helicopter?

d= ____ m

(c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at 1.10 m/s?

v = _____ m/s

d = ___ m

Answer 1

A small mailbag is released from a helicopter that is descending steadily at 1.10 m/s.

(a) After 5.00 s, what is the speed of the mailbag?

v = ___m/s

v(f) = v(i) + a*t. [v(f) is final velocity and v(i) is initial velocity]

Here a would. be gravitational acceleration which is 9.8 m/s²

v(initial)=1.10 m/s

After 5 seconds

v(final)=1.10 + 5*(9.8) = 50.1 m/s in the downward direction

(b) How far is it below the helicopter?

d= ____ m

Distance (d) that bag drops would be d=v(i)*t + (1/2)*a*t^2.

d =1.10*(5)+ (1/2)*(9.8m/s²)(5s)² which is 5.50 + 122.5 m = 128 m

In 5 seconds, helicopter would descend (or drop) by 5 * 1.10 = 5.50 meters therefore bag would be 122.5 meters below the helicopter

(c) What are your answers to parts (a) and (b) if the helicopter is rising steadily at 1.10 m/s?

to solve (c), just change v(i) in the equations above to -1.10 m/s therefore

v(f). = -1.10 + 9.8 *5 = 47.9 m/s in downward direction

Answer 2

To answer part a)

v(final)=v(initial) + a*t

Let a=-9.8 m/s²

v(initial)=0 m/s

after 5 seconds,

v(final)=0 + 5*(-9.8) or -49 m/s²

For part b)

d=v(initial)*t + (1/2)*a*t²

Distance of bag=0*(5)+ (1/2)*(-9.8m/s²)(5s)² or -122 m.

DIstance of helicopter =-1.10m/s* 5s or -5.50 m

Mailbag is -122 -(-5.50) or -116.50 lower than the helicopter