Learn formula for the sum of the roots of a quadratic equation and equation of trajectory, practice graphing quadratic and angle made by trajectory, maximum height of a projectile.
\(H_{max} =\dfrac{u^2 sin^2\theta}{2g}\)
\(y_0 \leq H_{max}\)
Case 1
If y_{0} < H_{max },_{ }then bullet can hit the plane at two points A and B, as the bird can attain altitude y_{0} at two points, as shown in figure.
Case 2
If y_{0 =} H_{ max }, only one point is possible where the bullet can hit the bird, as shown in figure.
\( y = x \,tan\, \theta - \dfrac{g\,x^2}{2\,u^2 cos^2\,\theta}\)
Above equation is a quadratic equation in terms of \(x\).
\(\left(\dfrac{g}{2\,u^2 cos^2\theta}\right)x^2 - (tan\,\theta)x + y =0\)
Comparing above equation with general quadratic equation
\(ax^2 + bx +c =0\)
\(a= \dfrac{g}{2\,u^2\,cos^2\theta}\) , \( b = –tan\,\theta,\,\,\,\,\, c = y\)
Discriminant is denoted by D and is equal to b^{2} - 4ac
\(D = b^2 – 4ac\)
A Three values
B One value
C Distinct two values
D No value
At maximum height, both the \(x-\) coordinates are equal i.e., there is only one value of \(x\) at maximum height.
\( QR = x_1 = OP\)
Range, \(R = x_2 + x_1\)
\(R = x_1 + x_2\) \([\because\,\,\, x_1 = x_2 = x]\)
\(R = 2x\)
\( y = x\, tan\,\theta - \dfrac{g\,x^2}{2\,u^2\,cos^2 \theta}\)
Above equation is a quadratic equation in \(x\).
On rearranging it,
\(x^2 - \dfrac{2\,u^2 x\,cos^2\theta}{g} tan\,\theta\,+ \dfrac{2\,u^2 \,cos^2\theta}{g} y =0\)
\(x^2 - \dfrac{\,u^2 \,sin\,2\theta}{g} \,x + \dfrac{2\,u^2 \,cos^2\theta}{g} y =0\) ....(i)
\(ax^2 + bx +c =0\)
\(a=1 \), \(b = \dfrac{- u^2 \, sin \,2 \theta}{g}\), \(c = \dfrac{2 \,u^2 \, cos^2 \theta}{g}y\)
Sum of roots = \(-\dfrac{b}{a}\)
Sum of roots = \(\dfrac{u^2\,sin\,2\theta}{g}\)
Sum of roots = Range
\(y_0 \leq H_{max}\)
Case 1:
Case 2 :
If y_{0 =} H_{ max }, only one point is possible where the bullet can hit the bird, as shown in figure.
\( y = x \,tan \,\theta - \dfrac{g\,x^2}{2\,u^2 cos^2\theta}\)
Above equation is a quadratic equation in terms of \(x\)
Arranging above equation
\(\left(\dfrac{g}{2\,u^2 cos^2\theta}\right)x^2 - (tan\,\theta)x + y =0\)
Comparing above equation with general quadratic equation
\(ax^2 + bx +c =0\)
\(a= \dfrac{g}{2\,u^2\,cos^2\theta}\) , \( b = –tan\,\theta,\,\,\,\,\, c = y\)
Discriminant is denoted by D and is equal to b^{2} - 4ac
\(D = b^2 – 4ac\)
Step 1 : Write the equation of trajectory.
Step 2 : Put the values of \( y, u ,\) \(\theta\) and \(g\).
Step 3 : Solve the quadratic equation for \(x\) and get two values of \(x\).
A 75 m, 15 m
B 5 m, 100 m
C 60 m, 20 m
D 150 m, 75 m
\(y= x \, tan\, \theta - \dfrac{g\,x^2}{2\,u^2cos^2\theta}\)
\(y= x \, tan\, \theta - \dfrac{g\,x^2}{2\,u^2 cos^2 \theta}\)
\(y_0= x_0 \, tan \,\theta - \dfrac{g\,x_0^2}{2\,u^2 cos^2 \theta}\)
\(y= x \, tan\, \theta - \dfrac{g\,x^2}{2\,u^2 cos^2 \theta}\)
\(y_0= x_0 \, tan \,\theta - \dfrac{g\,x_0^2}{2\,u^2 cos^2 \theta}\)
Step 1: Write the equation of the trajectory of the particle.
Step 2: Put the values of \(x,\, y,\,g\,\) and \(u\).
Step 3 : Solve for \(\theta\) and get two values of \(\theta\).
A 70º and 30º
B 30º and 60º
C 45º and 65º
D 45º and 85º