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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 18 m/s at an angle of 74° 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 between the two climbers?

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Depends on how long it has been travelling. If it is horizontal when it's been caught then ot must have reached zero velocity. No matter the angle gravity works at 1m/s/s. Do the math.

Posted on Sep 18, 2009

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This is a right angle triangle problem, and is way easier if you draw a picture. One climber is throwing a bag to another climber above and off to the side. The second climber is at a 74 degree incline from the first, 90 degrees would be looking straight up.

When the first climber lets go of the bag, its total velocity is 18m/s. Part of this velocity is upward, part is to the side. Looking at the angles, each the upward and side velocities would be equal if the pack was released from 45 degrees. Being as it was released from 74 degrees, much closer to vertical, the upward velocity component will be bigger.

sin(74)=0.9613, or in english the vertical component will be 96.13% of the total velocity.

Initial upward velocity = 18m/s*0.9613 = 17.3m/s

Total time of flight = 17.3m/s / 9.8m/s/s (deceleration due to gravity) = 1.7656seconds

Initial horizontal velocity = 18*cos(74) = 4.96m/s

Because there is no, or negligible, loss in horizontal velocity due to external forces:
Horizontal distance between climbers = 1.7656seconds*4.96m/s = 8.76 meters.

Posted on Sep 19, 2009

  • Nathan Sep 19, 2009

    The last missing part of the above solution is the kinematics equation:

    d = vi*t + 1/2*a*t^2

    = 17.3*1.7656 + 0.5*9.8*(1.7656)^2 = 45.82 meters

  • Nathan Sep 19, 2009

    The 9.8 in the last equation should be negative, so it comes out to
    15.27 meters vertical between the two. Sorry for the confusion.



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