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

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here2help

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 = -2gh, ( -ve sign because gravity acts downward)
0 - (20²) = -29.81h, 19.62h = 400, 
h = 400/19.62 = 20.38m.

Cricetus

At "20.38 m" height, the object will be rise. A further explanation is below..

Given:

Initial Kinetic energy,

  • 400 J

Mass,

  • 2 kg

As we know the formula,

→ \(Initial \ K.E =\frac{mass\times velocity^2}{2}\)

→               \(400=\frac{2\times V^2}{2 }\)

                 \(V^2 = 400\)

                  \(V = 20 \ m/sec\)

hence,

→ \(Vf^2-V^2 =-2gh\)

   \(0-(20)^2=-2\times 9.81\times h\)

   \(19.62\times h = 400\)

               \(h = \frac{400}{19.62}\)

                  \(= 20.38 \ m\)

Thus the above answer is right.

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