## Velocity Height Calculation check

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

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 Quote by Kev1n 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

 Quote by Kev1n 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?

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