A 2-kg object is thrown vertically upward with an initial kinetic energy of 400 joules. To what height will the object rise?

At "20.38 m" height, the object will be rise. A further explanation is below..Given:Initial Kinetic energy,400 JMass,2 kgAs we know the formula,→ [tex]Initial \ K.E =\frac{mass\times velocity^2}{2}[/tex]→               [tex]400=\frac{2\times V^2}{2 }[/tex]                  [tex]V^2 = 400[/tex]                   [tex]V = 20 \ m/sec[/tex]hence,→ [tex]Vf^2-V^2 =-2gh[/tex]    [tex]0-(20)^2=-2\times 9.81\times h[/tex]    [tex]19.62\times h = 400[/tex]                [tex]h = \frac{400}{19.62}[/tex]                   [tex]= 20.38 \ m[/tex]Thus the above answer is right. Learn more:https://brainly.com/question/3678943

initial kinetic energy = 400J,initial kinetic energy = [mass * Velocity²]/2 = [m * V²]/2,[2 * V²]/2 = 400, V² = 400, V = √400 = 20 m/sec,this V is initial velocity of object & it will keep on decreasing due to opposing gravitational force. At one point this will become zero & after that object will start to fall downward. If h is the maximum height it can reach, g is acceleration due to gravity & Vf if final velocity, then we can write, Vf² - V² = -2*g*h, ( -ve sign because gravity acts downward)0 - (20²) = -2*9.81*h, 19.62*h = 400, h = 400/19.62 = 20.38m.