Answer:
His speed when he leaves the ground 4.47 m/s.
Explanation:
Kevin's potential energy at the highest point of the jump is 1000 joules. Let us assume that Kevin jumps straight up in the air to a height of 1 meter.
It is required to find the Kevin's speed when he leaves the ground It is based on the conservation of energy. When it leaves the ground, the potential energy gets converted to its kinetic energy such that,
[tex]E=1000\ J\\\\mgh=1000\\\\m=\dfrac{1000}{10\times 1}=100\ kg[/tex]
Kinetic energy is given by :
[tex]E=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2E}{m}} \\\\v=\sqrt{\dfrac{2\times 1000}{100}} \\\\v=4.47\ m/s[/tex]
So, his speed when he leaves the ground 4.47 m/s.
