A body is thrown up from the ground with a velocity 39.2 m/s. At what height will its kinetic energy be reduced to one fourth of its original kinetic energy?

we know that

E' = (1/4)E

so,

(1/2)mv² = (1/4).[(1/2)mu²]

or

v' = (1/2)u = (1/2)x39.2

so,

v' = 19.6 m/s

also

v² - u² = 2as

here

u = 39.2 m/s

v = 19.6 m/s

a= -9.8 m/s²

so, height

s = (v² - u²) / 2a

s = (384.16 - 1536.64) / (2x-9.81)

thus, we get

s = 58.74 m