# [Mechanics] How to find speed of projection Watch

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This is the question:

I have done part (a) and (b) and got

(a) 35.3J

(b) 16.5J

This is my workings:

I am not sure how to use the work done and change in kinetic energy (work-energy) principle to solve for the projection speed. Currently I think the projection speed is the speed at which the particle was moved at up the slope when it came to instantaneous rest 3m up then gravity being the resultant force brought the particle back down but I don’t know how to manipulate and find this projection speed. Any intuitive help would be appreciated in understanding this, Thanks.

Could you use SUVAT on this?

Is it the work-energy principle?

I have done part (a) and (b) and got

(a) 35.3J

(b) 16.5J

This is my workings:

I am not sure how to use the work done and change in kinetic energy (work-energy) principle to solve for the projection speed. Currently I think the projection speed is the speed at which the particle was moved at up the slope when it came to instantaneous rest 3m up then gravity being the resultant force brought the particle back down but I don’t know how to manipulate and find this projection speed. Any intuitive help would be appreciated in understanding this, Thanks.

Could you use SUVAT on this?

Is it the work-energy principle?

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(Original post by

This is the question:

I have done part (a) and (b) and got

(a) 35.3J

(b) 16.5J

This is my workings:

I am not sure how to use the work done and change in kinetic energy (work-energy) principle to solve for the projection speed. Currently I think the projection speed is the speed at which the particle was moved at up the slope when it came to instantaneous rest 3m up then gravity being the resultant force brought the particle back down but I don’t know how to manipulate and find this projection speed. Any intuitive help would be appreciated in understanding this, Thanks.

Could you use SUVAT on this?

Is it the work-energy principle?

**BrandonS15**)This is the question:

I have done part (a) and (b) and got

(a) 35.3J

(b) 16.5J

This is my workings:

I am not sure how to use the work done and change in kinetic energy (work-energy) principle to solve for the projection speed. Currently I think the projection speed is the speed at which the particle was moved at up the slope when it came to instantaneous rest 3m up then gravity being the resultant force brought the particle back down but I don’t know how to manipulate and find this projection speed. Any intuitive help would be appreciated in understanding this, Thanks.

Could you use SUVAT on this?

Is it the work-energy principle?

The particle is projected with some speed v and initially it has kinetic energy equal to (1/2)mv^2. This is the total energy of the particle.

As it moves up, two things happen:

(A) since the plane is rough, there is friction. Friction is not a conservative force, therefore the particle loses a certain amount of energy due to friction,

(B) since the particle is moving up, it gains height with respect to its original position. Therefore the remaining KE is being slowly conserved and turned into GPE.

SO when it reaches the max height it can and stops, thats when it has lost some energy due to friction and the remaining energy has been completely converted into GPE.

NOTE that the total energy lost and GPE you found in the first two parts of the question must add up to the total initial energy.

Work out what this total energy is and solve the equation (1/2)mv^2 = (total energy) for the speed of projection v.

Last edited by RDKGames; 4 weeks ago

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(Original post by

Energy conservation is all you need. (Or rather, work-energy principle as you have it)

The particle is projected with some speed v and initially it has kinetic energy equal to (1/2)mv^2. This is the total energy of the particle.

As it moves up, two things happen:

(A) since the plane is rough, there is friction. Friction is not a conservative force, therefore the particle loses a certain amount of energy due to friction,

(B) since the particle is moving up, it gains height with respect to its original position. Therefore the remaining KE is being slowly conserved and turned into GPE.

SO when it reaches the max height it can and stops, thats when it has lost some energy due to friction and the remaining energy has been completely converted into GPE.

NOTE that the total energy lost and GPE you found in the first two parts of the question must add up to the total initial energy.

Work out what this total energy is and solve the equation (1/2)mv^2 = (total energy) for the speed of projection v.

**RDKGames**)Energy conservation is all you need. (Or rather, work-energy principle as you have it)

The particle is projected with some speed v and initially it has kinetic energy equal to (1/2)mv^2. This is the total energy of the particle.

As it moves up, two things happen:

(A) since the plane is rough, there is friction. Friction is not a conservative force, therefore the particle loses a certain amount of energy due to friction,

(B) since the particle is moving up, it gains height with respect to its original position. Therefore the remaining KE is being slowly conserved and turned into GPE.

SO when it reaches the max height it can and stops, thats when it has lost some energy due to friction and the remaining energy has been completely converted into GPE.

NOTE that the total energy lost and GPE you found in the first two parts of the question must add up to the total initial energy.

Work out what this total energy is and solve the equation (1/2)mv^2 = (total energy) for the speed of projection v.

Edit: It won’t let me rep you because I need to rep someone else first

Last edited by BrandonS15; 4 weeks ago

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