Solve Work-Energy Problem: Stone Thrown at Mango Tree

[Dumbledore211]

Homework Statement

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.

Homework Equations

v^2=u^2-2gh

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..

 

[CWatters]

If the boy uses half of the energy.. ??

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

 

[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...

 

[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

 

[HalfLostAndSearching]

I would like to commend your attempt at solving this work-energy problem. Your solution seems reasonable and follows the correct equations. However, I would like to point out a few things that can help improve your solution.

Firstly, when solving a problem like this, it is important to clearly define your variables and units. In this case, it would be helpful to state the initial and final velocities as v1 and v2, respectively, and specify their units as m/s. This will make it easier for the reader to understand your solution.

Secondly, when using equations, it is important to show all the steps in your calculation, rather than just stating the final equation. This will help to avoid any errors and make it easier for others to follow your solution.

Lastly, I would suggest double-checking your units in your final answer. The final height should have units of meters, not meters per second. So the correct answer would be 4.9 meters.

Overall, your solution seems to be correct and well thought out. Keep up the good work!