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

Answer:

 h = 5.071 m

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

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

$0 = 9.97^2 - 2\times 9.8 \times h$

  19.6 h = 99.4

   h = 5.071 m

hence, vertical height between the two climbers is 5.071 m