a basketball player can leap upward 0.5 m. what is his initial
velocity at the start of his leap and how long does the basketball
player remain in the air?

hi again shellyqu!!


Please recall the kinematics equations that can be used for this problem:
                    
s, v0, a and t   ----->     s = v0*t + 1/2*a*t^2 
vf, v0, a and s  ----->  vf^2 = v0^2 + 2*a*s 
vf, v0, a and t  ----->    vf = v0 + a*t  
s, v0, vf and t  ----->     s = 1/2*(v0 + vf )*t


what is his initial velocity at the start of his leap?

We know the following:
s = 0.5 m
vf = 0 m/s
a = g = 9.81 m/s^2
v0 = unknown

We can use the following equation:
vf, v0, a and s  ----->  vf^2 = v0^2 + 2*a*s 

==> v0^2 = vf^2 - 2*a*s = 
         = 0 m^2/s^2 - 2 * (-9.81 m/s^2) * 0.5 m =
         = 9.81 m^2/s^2
Note: I used a = -9.81 because the acceleration is negative (it is
against the motion).

==> v0 = sqrt(9.81 m^2/s^2) = 3.132 m/s

His initial vertical speed is 3.132 m/s .

                 ---------------------

how long does the basketball player remain in the air?

Since we know the initial speed we can use the following equation:
vf, v0, a and t  ----->    vf = v0 + a*t

==> t = (vf - v0) / a =
      = (0 m/s - 3.132 m/s) / (-9.81 m/s^2) =    
      = (-3.132) / (-9.81)  s =
      = 0.319 seg
Note: I used a = -9.81 because the acceleration is negative (it is
against the motion).

But this is the time until he reach the maximum height, we need the
total time. With no air resistance, the player will spend an equal
amount of time rising to the top as he spends falling from the top to
the ground. Since we have already found the half time of flight, we
need only to double that value to get the total time of flight, so the
total time is:

T = 2*0.319 seg = 0.638 seg.

-------------------------------------------------

I hope that this helps you. Feel free to request for a clarification
if you need it.

Regards.
livioflores-ga

Not much of a basketball player...

depends on his weight(gravity-vise) and much on his muscle
strength(think of it as a spring)

Think of it as 'drop a basketball player from 0.5m'

How long until he hits the ground ?
Then double it.

At what speed does he hit the ground ?

You should be able this question by considering what would happen if
you took a solid object and dropped from a height of .5 meters.  The
key is to realize that the players trajectory on the way up is exactly
the same as it is on the way down.

Doesn't d=.5at^2?  If it does, then the time for the player to fall
the distance of .5 meters is about .316 seconds and the player's
velocity when he lands is about 3.16 m/s (doesn't v=at?)

Therefore, the player's initial velocity when he leaps up is 3.16 m/s
and he takes .316 seconds to reach the apex of his jump (which means
he spends a total of .632 seconds in the air)

it all depends i guess on where you round too. if you do keep the
constants(i.e. keep gravity as g and make no assumption then the time
in the air is equal to 2*sqrt(g) which is 0.639s to 3 decimal places.