A small first-aid kit is dropped by a rock climber who is descending steadily at 1.3 m/s. After 2.5 s, what is the velocity of the first-aid kit, and how far is the kit below the climber?

D = 30.625 m

Explanation:

given,

Speed of the climber = 1.3 m/s

time = 2.5 s

acceleration due to gravity = 9.8 m/s²

initial speed of the kit = 1.3 m/s

velocity of the kit after 2.5 s

using equation of motion

v = u + a t

v = 1.3 + 9.8 x 2.5

v = 25.8 m/s

distance travel by the kit in 2.5 s

v² = u² + 2 g h

25.8² = 1.3² + 2 x 9.8 x h

19.6 h = 663.95

h = 33.875 m

distance travel by the rock climber in 2.5 s

distance = speed of climber x time

h' = 1.3 x 2.5

h' = 3.25 m

Distance between kit and rock climber

D = h - h'

D = 33.875 - 3.25

D = 30.625 m

The kit is 30.625 m below climber.