Question

A rock climber throws a small first aid kit to another climber who is higher on the mountain. The initial velocity of the kid is 11 m/s at an angle of 65°. The instant the kit is caught, it is traveling horizontally. What is the vertical height between the two climbers?

Answer 1

h = 5.071 m

Step-by-step explanation:

given,

initial velocity of the kit, v = 11 m/s

angle with horizontal = 65°

vertical height between the two climbers= ?

initial vertical velocity,

v_i = v sin θ

v_i = 11 sin 65° = 9.97 m/s

final vertical velocity = 0 m/s

using equation of motion

`v_f^2 = v_i^2 + 2 g h`

`0 = 9.97^2 - 2* 9.8 * h`

19.6 h = 99.4

h = 5.071 m

hence, vertical height between the two climbers is 5.071 m