Another Physics Problem, please help!!! ='(?

Question

A small first-aid kit is dropped by a rock climber who is descending steadily at 2.2 m/s.

a. After 4.1 s, what is the velocity of the first-aid kit? Answer in units of m/s.

b. How far is the kit below the climber after the 4.1 s? Answer in units of m.

Answer 1

<< After 4.1 s, what is the velocity of the first-aid kit? Answer in units of m/s. >>

Working formula is

Vf - Vo = gT

where

Vf = final velocity = velocity of kit after 4.1 sec.

Vo = initial velocity = 2.2 m/sec

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

T = time = 4.1 s (given)

Substituting appropriate values,

Vf - 2.2 = 9.8(4.1)

Vf = 42.38 m/sec.

<< How far is the kit below the climber after the 4.1 s? Answer in units of m. >>

After 4.1 sec., the climber has descended a distance = 2.2*4.1 = 9.02 meters

and the kit has dropped,

Y = VoT + (1/2)(gT^2)

where

Y = distance that kit has fallen after 4.1 sec.

and all the other terms have been previously defined.

Substituting values,

Y = 2.2(4.1) + (1/2)(9.8)(4.1)^2

Y = 9.02 + 82.37 = 91.39 meters

Let

D = distance between kit and climber after 4.1 sec.

D = 91.39 - 9.02

D = 82.37 meters

Hope this helps.