A 0.155 kg arrow is shot upward at 31.4 m/s. What is its kinetic
energy (KE) when it leaves the
bow?

2.43J

Explanation:

Given parameters:

Mass of the arrow = 0.155kg

Velocity = 31.4m /s

Unknown:

Kinetic energy when it leaves the bow = ?

Solution:

The kinetic energy of a body is the energy in motion of the body;

 it can be derived using the expression below:

   K.E  =  m v²

m is the mass

v is the velocity

 Solve for K.E;

    K.E  =    x 0.155 x 31.4 = 2.43J

the entire statement is true and correct.  

there's no diagram, nothing is illustrated, and no question is asked.

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