Velocity Height Calculation check


by Kev1n
Tags: calculation, check, height, velocity
Kev1n
Kev1n is offline
#1
Oct25-09, 09:33 AM
P: 39
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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Andrew Mason
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#2
Oct25-09, 10:06 AM
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Quote Quote by Kev1n View Post
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
Kev1n is offline
#3
Oct27-09, 06:31 AM
P: 39
Quote Quote by Kev1n View Post
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?

Andrew Mason
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#4
Oct27-09, 10:20 AM
Sci Advisor
HW Helper
P: 6,559

Velocity Height Calculation check


Quote Quote by Kev1n View Post
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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