Question

An object with a mass of 6kg is projected vertically upward and reaches a maximum height of 250m. Calculate the kinetic energy and velocity when the object is 200m above the ground while moving downwards.

Answer 1

What is the final potential energy? This will be the sum of the potential and kinetic energy at all points in time, including when it is 200m above the ground.

Answer 2

the potential energy at the max height is 14700J

Answer 3

And what is the potential energy when it is 200m above the ground?

Answer 4

11760J

Answer 5

So apply energy conservation \(U_{i}=U_f+K_f\), and \(K=\dfrac{mv^2}{2}\).

Answer 6

Yeah, energy conservation... Potential energy is just P = mgh, where m is mass, h is height and g is free fall acceleration. Let g = 10 m/s^2 then at height 250 m we have 6 x 10 x 250 = 15000 (J). It's total energy cause at this point no any motion, so kinetic energy is zero. At height 200 m we obtain 6 x 10 x 200 = 12000 (J) for potential energy, so kinetic energy must increase to 15000 - 12000 = 3000 (J). Then, due to mv^2/2, here is answer for velocity sqrt( 3000 x 2 / 6) = 10 m/s. So the answer is 10 m/s or -10 m/s depending on z axes direction (upward or downward).

Answer 7

i got that kinetic energy when my book asked me to calculate the kinetic energy when the object reaches a height of 200m while moving UPWARDS. in my book the answer for when moving downwards is 300J

Answer 8

the velocity in my book is 10m/s1though.

Answer 9

an object that weighs 6kg is dropped from 250m, what will its velocity be when it reaches 200m?

Answer 10

If you dig out why it isn't 31.62 m/s let me know.

Answer 11

will do. im stuck

Answer 12

total energy is conserved.
ke +pe is always constant .
so initialy when the object was to be thrown the pe is 0 and ke is maximum
at the maximum height net vel is 0
so ke =0
now net initial enegy=net final energy
(1/2 m )v ^2=mg*250
initial velocity v =root of 2 *g *250

potential energy +kinetic energy (at height of 200 m)=net potential energy at 250 m
ke=mg*250-mg*200
=m(2500-2000)
=6(500)
=3000 j

Answer 13

I just a little googled about parashutes. There are emergent ones with speed of descent up to 8 m/s, so this speed is relatively safe. If it's true, the conclusion i made is if the book is right we are all must be spidermen... at least.