Velocity Height Calculation check

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Homework Help Overview

The discussion revolves around a physics problem involving an object of mass 1.2 kg fired vertically from a barrel with a force of 25 N applied over 0.25 seconds. Participants are tasked with calculating the velocity upon leaving the gun, the height reached, and the time taken to return to the initial height.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the calculations for acceleration, velocity, height, and time using basic kinematic equations. Some question the accuracy of the velocity calculation, suggesting that the vertical firing aspect may not have been adequately considered. Others propose the need for a free body diagram to clarify the forces acting on the object.

Discussion Status

The discussion is ongoing, with participants providing calculations and questioning the assumptions made in the original approach. Guidance has been offered regarding the need to consider the net forces and acceleration in the context of vertical motion.

Contextual Notes

There is a mention of the need for a free body diagram to analyze the forces acting on the object, indicating that the problem may involve complexities not fully addressed in the initial calculations.

Kev1n
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1. An object of mass 1.2kg is fired vertically in a barrel velocity 25n acting over a period of 0.25 seconds, calculate A) The velocity when leave the gun. B). Height reached. C). Time taken to return to initial height.



2. a= f/m, V^2=U^2+2as



3. A), A=F/M for acceleration, 25/1.2= 20.8
Velocity = V0+AT = 0+20.8x0.25=5.2m/s

B). Height reached = V^2=U^2 + 2AS, 0=5.2^2 + 2 (-9.81)s
0= 27 - 19.62 x s, S = 27 / 19.62 = 1.38m

C). Time, S = (T/2( (UxV), T= 1.38 x 2/5.2 = 0.5
as require to arrive back = 2 x 0.5 = 1 Second

any comments appreciated
 
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Kev1n said:
1. An object of mass 1.2kg is fired vertically in a barrel velocity 25n acting over a period of 0.25 seconds, calculate A) The velocity when leave the gun. B). Height reached. C). Time taken to return to initial height.



2. a= f/m, V^2=U^2+2as



3. A), A=F/M for acceleration, 25/1.2= 20.8
Velocity = V0+AT = 0+20.8x0.25=5.2m/s

B). Height reached = V^2=U^2 + 2AS, 0=5.2^2 + 2 (-9.81)s
0= 27 - 19.62 x s, S = 27 / 19.62 = 1.38m

C). Time, S = (T/2( (UxV), T= 1.38 x 2/5.2 = 0.5
as require to arrive back = 2 x 0.5 = 1 Second

any comments appreciated
The velocity leaving the barrel is not correct.

In A you are not taking into account the fact that it is fired vertically. Do a free body diagram showing the forces acting, including the 25 N applied force.

AM
 
Kev1n said:
1. An object of mass 1.2kg is fired vertically in a barrel velocity 25n acting over a period of 0.25 seconds, calculate A) The velocity when leave the gun. B). Height reached. C). Time taken to return to initial height.



2. a= f/m, V^2=U^2+2as



3. A), A=F/M for acceleration, 25/1.2= 20.8
Velocity = V0+AT = 0+20.8x0.25=5.2m/s

B). Height reached = V^2=U^2 + 2AS, 0=5.2^2 + 2 (-9.81)s
0= 27 - 19.62 x s, S = 27 / 19.62 = 1.38m

C). Time, S = (T/2( (UxV), T= 1.38 x 2/5.2 = 0.5
as require to arrive back = 2 x 0.5 = 1 Second

any comments appreciated

Am I correct in understanding that
V0+at
Should be V0-Gt?
 
Kev1n said:
Am I correct in understanding that
V0+at
Should be V0-Gt?
No. Do a free body diagram. What forces act on the object? What is the net force? What is the acceleration? Use that value for "a" in the equation v = v0 + at.

AM
 

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