Question

If a rock climber accidentally drops a 77.5-g piton from a height of 215 meters, what would its speed be just before striking the ground? ignore the effects of air resistance.

Answer 1

Let us list out what we know from the question.

Initial Velocity
`V_(i)  = 0` since the piton is 'dropped'.

Vertical Displacement of the piton D = 215 m

Acceleration due to gravity
`a = 9.8 m/s^(2)`

Final Velocity
`V_(f) = ?`

Using the equation,
`V^(2) _(f) = V^(2) _(i)  + 2aD` and plugging in the known values, we get

`V^(2) _(f)  = 0^(2)  + 2(9.8)(215)`

Simplifying by taking square-root on both sides gives us
`V_(f) =  64.915 m/s`

Thus, the speed of the piton just before striking the ground is 65 m/s.