# Homework Help: Work-Energy Problem

1. Dec 24, 2013

### Dumbledore211

1. The problem statement, all variables and given/known data

A stone is thrown keeping target to a mango hanging in the branch of a mango tree. The velocity of the stone while hitting the mango is 9.8ms^-1. If the boy uses half of the energy used before the stone can reach the same height of the mango. Mass of the mango is 250gm.

2. Relevant equations
v^2=u^2-2gh

3. The attempt at a solution
According to the question the final velocity of the stone hitting the mango should be 9.8ms^-1. The velocity with which the boy projected the stone vertically should be greater than 9.8ms^-1 since it should be enough to counteract the gravitational acceleration. So we can say that the kinetic energy the the stone gains is half the kinetic energy the boy uses to project the stone. The workout is shown below and if I make any silly mistakes or misinterpret the question please point out my mistake and please see if the answer I got is reasonable enough
Let the velocity with which the boy projects the stone be v1
final velocity of the stone v2=9.8ms^-1
According to first condition
Ek1/2=Ek2
v1^2/2= v^2
v1= √2 v2
Hence, v1= 13.86ms^-1≈

v2^2=v1^2-2gh or, (9.8)^2= (13.86)^2-19.6h or, 96.04-192.1= -19.6h or, h= 96.06/19.6
h= 4.9meters
Sorry, I don't have the answer to this prob in my book so I have to make sure if this is the correct one..

2. Dec 24, 2013

### CWatters

The question doesn't make much sense. Can you check the wording.

3. Dec 24, 2013

### Dumbledore211

That is the part which I found bewildering to understand but I think what the question meant is that the final kinetic energy of the stone hitting the mango becomes half the kinetic energy of the boy that he uses to project the stone vertically . Then the whole problem makes sense.........

Last edited: Dec 24, 2013
4. Dec 24, 2013

### CWatters

...if you assume that the weight of the Mango is given just to confuse you.

Assuming you are correct then another way to solve the problem is to apply conservation of energy... the PE gain is where the other half of the initial energy went so.

mgh = 0.5 * 0.5mV12

then substitute V1 with √2 V2