Time of flight of a projectile

[CmbkG]

Homework Statement

A projectile of mass m is fired from the ground with a speed of 200m/s at an angle to the ground of 30 degrees. Assuming that the air resistance is linear, given by -mλv where v is the projectile's velocity and λ=0.1 s^-1, then find (taking g=10m.s^-2)

(a)the time of flight (time till it hits the ground again)
(b)the horizontal distance travelled
(c)the maximum height reached during the motion

Homework Equations

The Attempt at a Solution

Vx = Vo cosα
Vy = Vo sinα - gt

X = Xo + Vocosαt
Y = Yo + V0sinαt - gt^2/2

Suppose the particle is launched from the origin:

Xo=Yo=0

therefore: X = Vo cos αt
Y = Vo sinαt - gt^2/2

Eliminating t: t=X/VoCosα

Substitute into Y: Y=Tanαx - gX^2/2VO^2(Cosα)^2

Hence: 0=Tan(30X) - 10X^2/2(200)^2(Cos(30))^2

=Tan(30X) - 10X^2/60000

 

[HallsofIvy]

Homework Statement

A projectile of mass m is fired from the ground with a speed of 200m/s at an angle to the ground of 30 degrees. Assuming that the air resistance is linear, given by -mλv where v is the projectile's velocity and λ=0.1 s^-1, then find (taking g=10m.s^-2)

(a)the time of flight (time till it hits the ground again)
(b)the horizontal distance travelled
(c)the maximum height reached during the motion

Homework Equations

The Attempt at a Solution

Vx = Vo cosα
Vy = Vo sinα - gt

X = Xo + Vocosαt
Y = Yo + V0sinαt - gt^2/2

Suppose the particle is launched from the origin:

Xo=Yo=0

therefore: X = Vo cos αt
Y = Vo sinαt - gt^2/2

Eliminating t: t=X/VoCosα

Substitute into Y: Y=Tanαx - gX^2/2VO^2(Cosα)^2

Hence: 0=Tan(30X) - 10X^2/2(200)^2(Cos(30))^2

=Tan(30X) - 10X^2/60000

The angle is "a" so the coefficients are cos a and sin a, "Tanax" means "Tan(a) x", not "Tan(ax)".