Question

a pole vaulter clears a crossbar and takes 1.0 s to fall from the apex of his flighft to the landing pit. what was his vertical velocity at the end of the second?

Answer 1

`9.81\; {\rm m\cdot s^(-1)}` downward (assuming that
`g = 9.81\; {\rm m\cdot s^(-2)}` and that air resistance is negligible.)

Step-by-step explanation:

Under the assumptions, the pole vaulter would be accelerating downward at
`a_(y) = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}` under the effects of gravity. Note that
`a_(y)` is negative because gravitational attraction points downward. To find the vertical velocity
`v_(y)` after
`t = 1.0\; {\rm s}`, add the change in velocity
`a_(y)\, t` to the initial velocity
`u_(y)` (which needs to be found):

`v_(y) = u_(y) + a_(y)\, t`.

The question mentioned that at the beginning of the second, the pole vaulter was at the apex of the flight. The vertical velocity at that moment be
`u_(y) = 0\; {\rm m\cdot s^(-1)}`. Otherwise, if vertical velocity was positive, the pole vaulter would have moved upward even higher moments after; if vertical velocity was negative, the pole vaulter would have been at a higher position moments before.

Substitute
`u_(y) = 0\; {\rm m\cdot s^(-1)}`,
`a_(y) = (-g) = (-9.81)\; {\rm m\cdot s^(-2)}`, and
`t = 1.0\; {\rm s}` into the expression for
`v_(y)`:

`\begin{aligned}v_(y) &= u_(y) + a_(y)\, t \\ &= (0\; {\rm m\cdot s^(-1)}) + ((-9.81)\; {\rm m\cdot s^(-2)})\, (1.0\; {\rm s}) \\ &= (-9.81)\; {\rm m\cdot s^(-1)}\end{aligned}`.

(The value of
`v_(y)` is negative because the pole vaulter would be moving downwards.)

In other words, the vertical velocity of the pole vaulter at the given moment would be
`9.81\; {\rm m\cdot s^(-1)}` downward.