Analyze the flight of a projectile moving under the influence of gravity (mg) an

Question

Analyze the flight of a projectile moving under the influence of gravity (mg) and linear, i.e., laminar- flow drag (fD = -bv, with b > 0 constant). Neglect Phase 1; assume the projectile is launched from the origin at speed v1 and angle theta1 above the horizontal. WARNING: Neither the results for the drag-free projectile nor those for the projectile with turbulent- flow drag can be applied directly to this problem; you must go back to basic principles.

Write the equations of motion for the velocity components vx and vy of the projectile.

Determine the velocity components vx(t) and vy(t) of the projectile, as functions of time.

Determine the position coordinates x(t) and y(t) of a projectile as functions of time

Note from asker: Please don't forget use lamiar flow drag in your solutions. Please show work, I want to be able to understand the mechanics behind the answer. Thank you so much to whoever assists.

Explanation / Answer

For the given problem, as mentioned in the question, we will solve the motion along the horizontal and vertical directions independently and then resolve to find the required relations.

It needs to be realised here that the projectile will be moving under the influence of two forces at all times, which are the gravitational pull and the air drag. Plus, for the motion along the horizontal, we will only have the air drag as the gravitational pull will only affect the motion along the vertical direction.

Here, Initial velocity along the horizontal = Ux = V1 cos

Also initial velocity along vertical = Uy = V1 Sin

Plusm the deceleration along the horizontal would be bv/m

Hence, dv/dt = -bv/m

or dv / v = -bdt/m

We integrate from initial to V to obtain the relation for V. That is, ln[V/V1Cos] = -bt/m

Or Vx(t) = V1 Cos * e-bt/m

Further, for motion along the vertical direction, acceleration = -bv/m - g

Hence, dv/dt = -(bv/m + g)

dv / (bv/m + g) = -dt

Integrating from 0 to t, we get:

ln[(bVy/m + g) / (bV1Sin + g)] = -bt/m

[(bVy/m + g) / (bV1Sin + g)] = e-bt/m

bVy/m =  (bV1Sin/m + g)e-bt/m - g

Vy =(m/b) [(bV1Sin/m + g)e-bt/m - g]

Now, we use the above relations for Vx and Vy to determine the expressions for x(t) or y(t)

dx/dt = Vx = V1 Cos * e-bt/m

dx = V1 Cos * e-bt/mdt

Integrating from t = 0 to t, we get the displacement along x as:

X(t) = (m/b)V1 Cos * [e-bt/m-1]

Again dy/dt = Vy = (m/b) [(bV1Sin/m + g)e-bt/m - g]

or dy = (m/b) [(bV1Sin/m + g)e-bt/m - g] dt

Integrating, we get the expression for y as:

Y(t) = (m/b) [(V1Sin + mg/b)(e-bt/m - 1 ) - gt]

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