Question

A body of mass 1kg drops from the top of tower of height 50m what will the kinetic energy 10m below the top

Answer 1

FIRST WE WILL FIND VELOCITY,

v²=u²+2gh [u=0]

v²=o² + 2×-10×-10 [g=-10,h=-10m]

v=40m/s

now, KE = ½MV²

-> ½×1×40×40

= 80 Joules

Answer 2

Answer:

96j

Explanation:

loss of p. E =.gain of k. E

mgh. =1/2 mv2

1*9.8*50

490

just below 10m it's p. E

is

1*9.8*40

392

thus loss of potential is equal to 490-392=98