(adsbygoogle = window.adsbygoogle || []).push({}); 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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# Velocity Height Calculation check

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