Question

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 13.0 m/s at an angle of 69.0∘∘ above the horizontal. At the instant when the kit is caught, it is traveling horizontally, so its vertical speed is zero. What is the vertical height difference between the two climbers?

Answer 1

Consider the vertical motion:

Initial vertical velocity = 13 sin69 = 12.1365 m/s

Vertical acceleration = - 9.8 m/s2

Final vertical velocity = 0 m/s (as finally, the kit is traveling horizontally)

Now, under uniform acceleration, v^2=u^2+2as, where v is final velocity, u is initial velocity, a is acceleration and s is displacement.

So, 0*0=12.1365*12.1365+2*(-9.8)*s

=>s=12.1365*12.1365/(2*9.8) = 7.5 m.

So, required vertical height difference = 7.5 meters.