a ball is dropped from a tall building with height 30.0m. What is the speed of t

Question

a ball is dropped from a tall building with height 30.0m. What is the speed of the ball at the height of 20.0m? What part of energy is in kinetic energy form at height of 20.0m? At what height does the ball have equal amount of gravitational potential and kinetic energy?

I solved the first part:

1/2mv^2 = mgy2 - mgy1 ==> v^2 = 2g(y2 - y1) ==>v^2 = sqrt of 196m/s ==> v = 14.0m/s

Can anyone help me with the latter two questions? I'm not sure how to set it up. I know how to find the total energy in the form of kinetic energy: K2/E1, but I don't think that helps me here.

Explanation / Answer

here,

height of the building , h = 30 m

final height of the ball , h' = 20 m

let the speed at that point be v

using conservation of energy

0.5 * m * v^2 = m * g * ( h - h')

0.5 * v^2 = 9.8 * ( 30 -20)

v = 14 m/s

the ball's velocity at 20 m height is 14 m/s

kinetic energy at height h = 20 m, KE = 0.5 * m * v^2/ m*g*h * 100

KE = 0.5 * 14^2 /( 9.8 * 30) * 100

KE = 33.33 %

kinetic energy is 33.33 % of the total energy

as total energy is some of the potential energy and the kinetic energy

and that remains constant

therefore ,

at height h = 15 m

the potential energy is equal to the kinetic energy

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