A 108 kg basketball player crouches down 0.400m while waiting to jump. After exerting a force on the floor through this 0.400m, his feet leave the floor and his center of gravity rises 0.924 m above its normal standing erect position. In kilonewtons, what average force did he exert on the floor? Be sure to include the force that supports his weight as well as the force needed to accelerate him upward.

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer

the inner core.

good luck: )

klicker

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west