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A 67 kg high jumper leaves the ground with a vertical velocity of 6.4 m/s.

How high can he jump?

The acceleration of gravity is 9.8 m/s².

Answer in units of meters (m).

Answer :

The height of the jump having a mass of 67kg with the vertical velocity of 6.4m/s is 2.089m.

What is Kinetic and Potential energy?

Potential energy is defined as the energy stored in an object or system by virtue of its position or arrangement of parts while kinetic energy is defined as the energy of the moving particles of an object or system.

Potential energy is represented as :

P.E.= mgh

Kinetic energy is represented as :

[tex]K.E.= 1/2mv^2[/tex]

where, m = mass of the object

h= height

v= velocity

g= acceleration due to gravity which is [tex]9.8m/s^2[/tex]

For above given example,

m= 67kg, v= 6.4m/s

we first find out Kinetic energy,

K.E.=[tex]\frac{1}{2} 67* (6.4)^2[/tex]= 2744.32/2= 1372.16 J

As we know, K.E.= P.E.

So, P.E.= 1372.16J = mgh

1372.16= 67* 9.8* h

height= 2.089 m

Thus, the height of the jump having a mass of 67kg with the vertical velocity of 6.4m/s is 2.089m.

Learn more about Kinetic energy, here:

https://brainly.com/question/26472013

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