# If a person glides through the air at a height of 215 m with a velocity of 25.4 m/s, suddenly they dive to a height of 178m. Whats the new velocity?

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The person is gliding at a height 215 m with a velocity 25.4 m/s. The sum of the kinetic energy and gravitational potential energy of the person is equal to (1/2)*m*v^2 + m*g*h where v is the velocity, m is the mass and h is the height of the person.

Initially the total energy is equal to (1/2)*m*25.4^2 + m*9.8*215

Let V represent the magnitude of the velocity of the person when the height is reduced to 178 m. The total energy of the person is (1/2)*m*V^2 + m*9.8*178. The law of conservation of energy states that the total energy of a closed system is constant. Here, the person can be considered to be a closed system. Equating the initial energy with the final energy gives:

(1/2)*m*25.4^2 + m*9.8*215 = (1/2)*m*V^2 + m*9.8*178

=> 25.4^2 + 4214 = V^2 + 3488.8

=> V^2 = 1370.36

=> `V ~~ 37.01`

**The magnitude of velocity of the person at 178 m is approximately 37.01 m/s.**