A rock climber throws a small first aid kit to another climber who is higher up the mountain. The initial velocity of the kit is 16 m/s at an angle of 55° above the horizontal. The kit is caught right at the peak of the its trajectory. What is the vertical height between the two climbers?

Answer 1

Vo = 16m/s[55o]
Yo = 16*sin55 = 13.11 m/s.

Y^2 = Yo^2 + 2g*h
h = (Y^2-Yo^2)/2g
h = (0-13.11^2)/-19.6 = 8.77 m.