Velocity Height Calculation check

In summary: 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.In summary, the object experiences a velocity of 5.2m/s when it leaves the barrel.
  • #1
Kev1n
40
0
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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  • #2
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
 
  • #3
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?
 
  • #4
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
 

1. What is "Velocity Height Calculation check"?

"Velocity Height Calculation check" is a scientific method used to calculate the velocity of an object in motion by measuring its height at two different points in time.

2. How does the "Velocity Height Calculation check" work?

The "Velocity Height Calculation check" works by using the formula v = ∆h/∆t, where v is the velocity, ∆h is the change in height, and ∆t is the change in time. This formula allows us to compare the height of an object at two different points in time and calculate its velocity.

3. What are the applications of "Velocity Height Calculation check"?

"Velocity Height Calculation check" is commonly used in physics and engineering to calculate the velocity of moving objects. It can also be used in sports to measure the speed of athletes, and in meteorology to track the movement of weather systems.

4. What are the limitations of "Velocity Height Calculation check"?

One limitation of "Velocity Height Calculation check" is that it assumes the motion of the object is constant and in a straight line. It also does not take into account external factors such as air resistance or friction, which can affect the accuracy of the calculated velocity.

5. How can the accuracy of "Velocity Height Calculation check" be improved?

The accuracy of "Velocity Height Calculation check" can be improved by using more precise measuring tools, such as laser sensors, and by taking multiple measurements at different points along the object's path. It is also important to consider external factors and make adjustments to the calculation if necessary.

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