Question

A branch falls from the top of a 95.0m tall redwood tree, starting from rest. Use work-kinetic energy theorem to determine the speed of the branch when it reaches the ground.

Answer 1

@Hero @Callisto

Answer 2

@ash2326 @chemENGINEER

Answer 3

@thomaster

Answer 4

work done = change in KE

apply it ?

Answer 5

Then it should be W = mV^2/2 - mVo^2/2

Answer 6

Your v0=0

Answer 7

Yup ! plug the values
and W = F.s

Answer 8

I have no initial velocity, hence. W= mV^2/2, is that right?

Answer 9

Correct

Answer 10

Then I will have mad = mV^2/2
ad = V^2/2, where a=g, right?
Hence V = sqrt(2gd)
V = sqrt(2(9.8)(95))
Resulting to V = 43.15m/s
Did I've done it correctly?

Answer 11

Yes

Answer 12

yup ! thats correct

Answer 13

you can verify it using the kinematics if u want,
until u get hang of conservation of energy concept

Answer 14

v^2 - u^2 = 2ad

put u = 0

you wil get the same equation

v = sqrt(2gd)

Answer 15

Yeah @ganeshie8 . Thanks >:)). I still have one more, can you guide me?

Answer 16

il try :)

Answer 17

Shot put competitions use metal balls with a mass of 16lb. A competitor throws the shut at an angle of 43.3 deg and releases it from a height of 1.82m above where it lands, and lands a horizontal distance of 17.7m from the point of release. What is the kinetic energy of the shot as it leaves the thrower's hand?