Michael jordan, el célebre basquetbolista, ganó el torneo de clavadas de la nba en 1988. para lograr la hazaña saltó 1.35 metros verticalmente ¿cuánto tiempo permaneció en los
20 cm de la parte más alta de su salto? ¿y en los 20 cm de su parte más baja?

(a) 0.40 s

First of all, let's find the initial speed at which Jordan jumps from the ground.

The maximum height is h = 1.35 m. We can use the following equation:

where

v = 0 is the velocity at the maximum height

u is the initial velocity

is the acceleration of gravity

Solving for u,

The time needed to reach the maximum height can now be found by using the equation

Solving for t,

Now we can find the velocity at which Jordan reaches a point 20 cm below the maximum height, so at a height of

h' = 1.35 - 0.20 = 1.15 m

Using again the equation

we find

And the corresponding time is

So the time to go from h' to h is

And since we have also to take into account the fall down (after Jordan reached the maximum height), which is symmetrical, we have to multiply this time by 2 to get the total time of permanence in the highest 20 cm of motion:

(b) 0.08 s

This part is easier since we need to calculate only the velocity at a height of h' = 0.20 m:

And the corresponding time is

So this is the time needed to go from h=0 to h=20 cm; again, we have to take into account the motion downwards, so we have to multiply this by 2:

graphing motion: mastery test

209be

what is the instantaneous acceleration of

the particle at point b?

0 meters/second?

10 meters/second?

12 meters/second?

velocity imetersisecond

15 meters/second?

20

time (seconds)

answer: 0 meters/second^2

what is the speed of a softball that travels 20m in 2secs

speed = 20/2 = 10 m/s