If the equation of projection is y=ax-bx^2, find initial velocity(u) and angle of projection(theta)?

2 Answers
Jun 25, 2017

Given that the equation of projection is y=ax-bx^2, we are to find initial velocity(u) and angle of projection(theta)

Horizontal component of velocity of projection is ucostheta

Vertical component of velocity of projection is usintheta

Let at the t th sec after its projection from origin its position be represented by the coordinates (x,y)

So x=ucosthetaxxt

=>t=x/(ucostheta).....[1]

Again

y=usinthetaxxt-1/2gxxt^2.....[2]

Combining [1] and[2] we get

y=usinthetaxxx/(ucostheta)-1/2gxx x^2/(u^2cos^2theta)

=>y=xtantheta- (gx^2)/(2u^2cos^2theta)....[3]

Comparing equation [3] with the given equation we get

a=tantheta

=>theta=tan^-1a

and

b=g/(2u^2cos^2theta)

=>u^2=g/(2b)sec^2theta

=>u^2=g/(2b)(1+tan^2theta)

=>u^2=g/(2b)(1+a^2)

=>u=sqrt(g/(2b)(1+a^2))

Jun 25, 2017

See below.

Explanation:

This is a typical kinematic problem. Assuming that the movement begins at the referential origin we have

{(x = v_x t),(y=v_y t-1/2 g t^2):}

where

t is the elapsed time
v_x the x-axis velocity component
v_y the y-axis velocity component
g the gravity acceleration.

Now the orbit equation is

y-ax+bx^2=0 substituting the parametric equations we have

1/2(g t + 2 a v_x - 2 b t v_x^2 - 2 v_y)t=0

This condition must be true for all t so

(g t + 2 a v_x - 2 b t v_x^2 - 2 v_y)=0, forall t

Now grouping coefficients

(g - 2 b v_x^2)t+2(a v_x - v_y)=0 so

{(g - 2 b v_x^2=0),(a v_x - v_y=0):}

Now solving for v_x, v_y

{(v_x=sqrt(g/(2b))),(v_y=a sqrt(g/(2b))):}

Now v=sqrt(v_x^2+v_y^2) = sqrt((g/2)((a^2+1)/b))

theta = arctan(v_y/v_x) =arctan(a)