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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?



Answer :

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 = -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.

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,

→ [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.

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