# Motion of a projectile with air resistance?

Please help.
A projectile is thrown verticali into the air with initial velocity of 20m per second (v0) at t=0
Using the quadratic model for air resistance and using a=dv/dt find t in terms of v, b, g and v0 and hence find the time taken for the projectile to reach the maximum hight.
b^2 = mg/0.2D^2
m =...
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Please help.

A projectile is thrown verticali into the air with initial velocity of 20m per second (v0) at t=0

Using the quadratic model for air resistance and using a=dv/dt find t in terms of v, b, g and v0 and hence find the time taken for the projectile to reach the maximum hight.

b^2 = mg/0.2D^2

m = 0.02kg, D = 0.03m, g = 9.81ms^-2, vo= 20ms^-1

In vector form, with origin at point of projection, I use

ma = -mg-(0.2D^2)(V^2)

mdv/dt = -mg-(0.2D^2)(V^2)

m/-mg-(0.2D^2)(V^2) dv = 1 dt

1/-g-k(V^2) dv = 1 dt , k=-(0.2D^2)/m

I intergrate and get

-1/2k(In(-g-k(V^2)) = t + A

Am I correct ?

If so wy cant I get the correct answer ? it sould be around 2 seconds.

Also using a = vdv/dx how do I find the hight.It should be 17.37m

Many thanks

A projectile is thrown verticali into the air with initial velocity of 20m per second (v0) at t=0

Using the quadratic model for air resistance and using a=dv/dt find t in terms of v, b, g and v0 and hence find the time taken for the projectile to reach the maximum hight.

b^2 = mg/0.2D^2

m = 0.02kg, D = 0.03m, g = 9.81ms^-2, vo= 20ms^-1

In vector form, with origin at point of projection, I use

ma = -mg-(0.2D^2)(V^2)

mdv/dt = -mg-(0.2D^2)(V^2)

m/-mg-(0.2D^2)(V^2) dv = 1 dt

1/-g-k(V^2) dv = 1 dt , k=-(0.2D^2)/m

I intergrate and get

-1/2k(In(-g-k(V^2)) = t + A

Am I correct ?

If so wy cant I get the correct answer ? it sould be around 2 seconds.

Also using a = vdv/dx how do I find the hight.It should be 17.37m

Many thanks

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