Velocity Height Calculation check

AI Thread Summary
An object with a mass of 1.2 kg is fired vertically with a force of 25 N over 0.25 seconds, leading to a calculated velocity of 5.2 m/s upon leaving the barrel. The height reached by the object is determined to be 1.38 meters using the kinematic equation. The total time taken for the object to return to its initial height is calculated to be 1 second. However, there are concerns regarding the accuracy of the velocity calculation, emphasizing the need for a free body diagram to account for forces acting on the object. Correct understanding of the equations of motion and forces is crucial for accurate 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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