In basketball, hang time is the time that both of your feet are off the ground during a jump. the equation for hang time is t=2(2h/32)^1/2, where t is the time in seconds, and h is the height of the jump, in feet.

player 1 had a hang time of 0.9 s. player 2 had a hang time of 0.8 s. to the nearest inch, how much higher did player 1 jump than player 2?
.
is it 8? because i solved the problem and player 1s jump height minus player 2 is 0.68 feet which is 8.16 inches?

the value of v.w= -12.

step-by-step explanation:

we are given three vector r,v and w as:

r = < 5, -5, -2> ; v = < 2, -8, -8> ; w = < -2, 6, -5>

we know that r,v and w could also be represented as:

r=5i-5j-2k

v=2i-8j-8k

and w= -2i+6j-5k

where i,j and k are the unit vectors of x,y and z-axis respectively.

also we know that it has the following property:

i.i=1 ; j.j=1; k.k=1

and i.=j.i=i.k=k.i=j.k=k.j=0

i.e. when we multiply two vectors or we find it's dot product we need to only multiply the components of x coordinates with each other; y with each other and z with each other.

hence, v.w= (2i-8j- -2i+6j-5k)

                  = 2×(-2)+(-8)×(6)+(-8)×(-5)

                  = -4-48+40

                  = -12

hence, v.w= -12

idk

step-by-step explanation: