Make a model to describe the path a bullet takes after being fired from a gun 3.5 m from the ground traveling?

[The bullet is fired 'mostly' upwards into the air from the numbers given]

Important point is assuming constant deceleration so we can use the "suvat" equations.

Note here I use vectors. If you're unfamiliar with them, basically a vector ( A , B )' just means A horizontally (x direction) and B vertically (y direction) and we usually assume that 'right' and 'up' are the positive directions, like a graph.

s = displacement at time t = ? m

u = initial velocity at time 0 = 400 m/s in an unknown direction = (400cosA, 400sinA)

v = velocity at time t = ? m/s

a = constant acceleration = (0, -g)' m/s

t = time = 4 s

initial displacement = (0, 3.5)' m

final displacement = (2200, 0)' m

total time = T = ? s

where g = acceleration due to gravity, approx 9.81 m/s^(2)

and A is the angle (between 0 and 90 degrees) made between initial direction and horizontal.

and we pick from a choice of common equations the one(s) to get what we want:

v=u+at

s=(1/2)(u+v)t

s=ut+(1/2)at^2

v^2=u^2+2as

a=(v-u)/t

Consider the total journey of the bullet first, so at time T.

s=ut+(1/2)at^2 taken horizontally gives

2200 = 400cosAT + (1/2)*0*T^2 = 400cosAT

T = (11/2cosA)

s=ut+(1/2)at^2 taken vertically gives

-3.5 = (1/2)(-9.81)T^2

T = sqrt(3.5*2/9.81) = 0.84472338...

so cosA = 0.84472338... *2/11 = 0.1535860...

and A = arccos(0.1535860...) = 81.165197... degrees

So now consider the system at 4 seconds

s=ut+(1/2)at^2 taken horizontally gives

s = 4*400cosA + 0.5*16*0 = 1600cos(81.165197...) = 1600*0.1535860... = 245.73771... m

s=ut+(1/2)at^2 taken vertically gives

s = 4*400sinA + 0.5*16*(-9.81) = 1600sin(81.165197...)+ 8*(-9.81) = 1 502.5364... m

so distance travelled is sqrt(245.73771...^2 + 1 502.5364...^2) = 1 522.4988... m

= 1522.50 m approx

quickly and livid Its that stutter step that romps me up like the Roadrunner spurting away for Wiley Cyote with that Acme Atomic contraption.that has no guidance and finally ends up over the cliff.