Asked by ????
A body of mass 3kg is dropped from a tower of height 250m. then its kinetic energy after 3 seconds will be
Answers
Answer
Who said it fell the whole 250 m?
Check that
v = -g t
h = 250 - 4.9 t^2
at t = 3 seconds
h = 250 - 4.9 *9 = 250 - 44.1
so h = 206 meters still after 3 sec
now for real:
Ke = 1/2 m v^2
v = -9.81 * 3 = - 29.43
Ke = (1/2) (3) (29.43)^2 = 1,299 Joules
Check that
v = -g t
h = 250 - 4.9 t^2
at t = 3 seconds
h = 250 - 4.9 *9 = 250 - 44.1
so h = 206 meters still after 3 sec
now for real:
Ke = 1/2 m v^2
v = -9.81 * 3 = - 29.43
Ke = (1/2) (3) (29.43)^2 = 1,299 Joules
Answer
The robo replies are driving me nuts !
Answered by
🌺🐱Catwolf901🐺🌺
?
Answered by
Bot
The kinetic energy of the body after 3 seconds will be equal to its gravitational potential energy at the start of the fall.
Gravitational potential energy = mgh
Gravitational potential energy = (3 kg)(9.8 m/s2)(250 m)
Gravitational potential energy = 72,000 Joules
Kinetic energy = 72,000 Joules
Gravitational potential energy = mgh
Gravitational potential energy = (3 kg)(9.8 m/s2)(250 m)
Gravitational potential energy = 72,000 Joules
Kinetic energy = 72,000 Joules
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