E-StoreA reverse problem: Finding projectile parameters from its equation
18 May 2007 19:30:58 IST
A reverse problem: Finding projectile parameters from its equation
Most of you have read projectiles and know how to find the equation of trajectory when the initial velocity, angle of projection are given.
Lets consider a reverse problem:
Given a general quadratic equation, lets find its initial velocity, angle of projection, time of flight, etc., etc, etc ....
Consider a quadratic equation
y = ax2+bx+c -------------(1)
The X axis is considered to be the horizontal plane, and the Y axis is vetial, positive up.
Now, for this equation to represent a projectile, it should be a parabola that opens downwards, ie, a<0
Also, it should have two real roots. For the case of a single real root, the equation represents a particle thrown vertically upwards.
So
= b2 - 4ac >0
Lets assume that the particle is projected from the X axis only.
Then, if c>0, the particle is projected from the left of the origin.
If c = 0, the particle is projected from origin.
If c< 0, the particle is projected from the right of the origin.
Differentiating the equation wrt x,
dy/dx = 2ax + b
Equating it to zero, we get the critical point
xc = -b/2a
d2 y/dx2 = 2a <0
Hence xc is a point of maxima.
H = ymax = y(xc ) = -/4a -----------------(2)
Differentiating the equation wrt time t,
dy/dt = 2ax(dx/dt) + b(dx/dt) ---------------(3)
Differentiating again wrt t
d2 y/dt2 = (2ax + b)(d2 x/dt2) + 2a(dx/dt)2 ---------------(4)
Now,
d2 y/dt2 = -g & d2 x/dt2 = 0
Substituting these values in (4)
-g = (2a)vx2
vx =
(-g/2a), which is a constant. -----------------(5)
So the X component of velocity remains constant.
From (3),
vy = (2ax+b)vx -------------------(6)
Let the roots of the equation (1) be
& 
such that 
<. Then x = is the projection point. Hence at the projection point,
ux = (-g/2a) &
uy = (2a+b)vx
Now asis a root of the equation (1)
= (-b+ )/2a
(2a+b) =
Hence at projection point,
ux = (-g/2a) &
uy =(-g/2a)
So initial speed,
u = ux2 + uy2
= (1+)(-g/2a) --------------------(7)
Angle of projection
= tan-1 (uy / ux) = tan-1 ------------------(8)
Time of flight
T = 2uy/g =(-/2ag)
Horizontal range
R = 2uy ux /g = -/a ------------------(9)
These derived equations provide a complete description of the projectile.
u = [(1+)(-g/2a)]
= tan-1
R = -/a
T =(-/2ag)
H = -/4a
Lets take an example to see these formulas in work
Let the equation be
y = -x2+5x-6
The constant term is -ve, hence the particle is projected from the +ve X axis
= 1
u = [(1+)(-g/2a)] = 10 m/s
= tan-1 = 45 deg
R = -/a = 1 m
T =(-/2ag) = 1/(2 5) s
H = -/4a = 1/4 m
Lets consider a reverse problem:
Given a general quadratic equation, lets find its initial velocity, angle of projection, time of flight, etc., etc, etc ....
Consider a quadratic equation
y = ax2+bx+c -------------(1)
The X axis is considered to be the horizontal plane, and the Y axis is vetial, positive up.
Now, for this equation to represent a projectile, it should be a parabola that opens downwards, ie, a<0
Also, it should have two real roots. For the case of a single real root, the equation represents a particle thrown vertically upwards.
So
= b2 - 4ac >0Lets assume that the particle is projected from the X axis only.
Then, if c>0, the particle is projected from the left of the origin.
If c = 0, the particle is projected from origin.
If c< 0, the particle is projected from the right of the origin.
Differentiating the equation wrt x,
dy/dx = 2ax + b
Equating it to zero, we get the critical point
xc = -b/2a
d2 y/dx2 = 2a <0
Hence xc is a point of maxima.
H = ymax = y(xc ) = -
/4a -----------------(2)Differentiating the equation wrt time t,
dy/dt = 2ax(dx/dt) + b(dx/dt) ---------------(3)
Differentiating again wrt t
d2 y/dt2 = (2ax + b)(d2 x/dt2) + 2a(dx/dt)2 ---------------(4)
Now,
d2 y/dt2 = -g & d2 x/dt2 = 0
Substituting these values in (4)
-g = (2a)vx2
vx =
(-g/2a), which is a constant. -----------------(5)So the X component of velocity remains constant.
From (3),
vy = (2ax+b)vx -------------------(6)
Let the roots of the equation (1) be
& such that <. Then x = is the projection point. Hence at the projection point,ux =
(-g/2a) &uy = (2a
+b)vxNow as
is a root of the equation (1)= (-b+ )/2a(2a
+b) = Hence at projection point,
ux =
(-g/2a) &uy =
(-g/2a)So initial speed,
u =
ux2 + uy2=
(1+)(-g/2a) --------------------(7)Angle of projection
= tan-1 (uy / ux) = tan-1 ------------------(8)Time of flight
T = 2uy/g =
(-/2ag)Horizontal range
R = 2uy ux /g = -
/a ------------------(9)These derived equations provide a complete description of the projectile.
u =
[(1+)(-g/2a)]= tan-1 R = -
/aT =
(-/2ag)H = -
/4aLets take an example to see these formulas in work
Let the equation be
y = -x2+5x-6
The constant term is -ve, hence the particle is projected from the +ve X axis
= 1u =
[(1+)(-g/2a)] = 10 m/s= tan-1 = 45 degR = -
/a = 1 mT =
(-/2ag) = 1/(2 5) sH = -
/4a = 1/4 m
Comments (2)
Anil Sahu
Blazing goIITian
Joined: 20 Apr 2007 19:50:01 IST
Posts: 387
18 May 2007 22:51:28 IST
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