Solution 19
Step 1: The Trajectory Equation
The trajectory of a projectile is given by: $$ y = x \tan\theta – \frac{gx^2}{2u^2 \cos^2\theta} $$ Using the identity $\frac{1}{\cos^2\theta} = 1 + \tan^2\theta$: $$ y = x \tan\theta – \frac{gx^2}{2u^2} (1 + \tan^2\theta) $$ Rearranging this into a quadratic equation in terms of $\tan\theta$: $$ \left( \frac{gx^2}{2u^2} \right) \tan^2\theta – x \tan\theta + \left( y + \frac{gx^2}{2u^2} \right) = 0 $$
The trajectory of a projectile is given by: $$ y = x \tan\theta – \frac{gx^2}{2u^2 \cos^2\theta} $$ Using the identity $\frac{1}{\cos^2\theta} = 1 + \tan^2\theta$: $$ y = x \tan\theta – \frac{gx^2}{2u^2} (1 + \tan^2\theta) $$ Rearranging this into a quadratic equation in terms of $\tan\theta$: $$ \left( \frac{gx^2}{2u^2} \right) \tan^2\theta – x \tan\theta + \left( y + \frac{gx^2}{2u^2} \right) = 0 $$
Step 2: Condition for Maximum Range
For a given vertical displacement $y$ (where $y = +h$ or $y = -h$), we want to find the maximum possible horizontal range $x$. The quadratic equation above must have real roots for $\tan\theta$ if the target $(x, y)$ is reachable. At the maximum possible range (the boundary of the reachable region), the discriminant of this quadratic equation must be zero. $$ D = b^2 – 4ac = 0 $$ $$ (-x)^2 – 4 \left( \frac{gx^2}{2u^2} \right) \left( y + \frac{gx^2}{2u^2} \right) = 0 $$ While solving this for $x$ gives the maximum range value, we specifically need the angle. When the discriminant is zero, the quadratic has a single unique root: $$ \tan\theta = \frac{-b}{2a} = \frac{x}{2 \left( \frac{gx^2}{2u^2} \right)} = \frac{u^2}{gx} $$
For a given vertical displacement $y$ (where $y = +h$ or $y = -h$), we want to find the maximum possible horizontal range $x$. The quadratic equation above must have real roots for $\tan\theta$ if the target $(x, y)$ is reachable. At the maximum possible range (the boundary of the reachable region), the discriminant of this quadratic equation must be zero. $$ D = b^2 – 4ac = 0 $$ $$ (-x)^2 – 4 \left( \frac{gx^2}{2u^2} \right) \left( y + \frac{gx^2}{2u^2} \right) = 0 $$ While solving this for $x$ gives the maximum range value, we specifically need the angle. When the discriminant is zero, the quadratic has a single unique root: $$ \tan\theta = \frac{-b}{2a} = \frac{x}{2 \left( \frac{gx^2}{2u^2} \right)} = \frac{u^2}{gx} $$
Step 3: Solving for Case (a) – Target at height $h$ ($y = +h$)
First, find $x_{max}$ from the discriminant equation: $$ x^2 = \frac{2gx^2}{u^2} \left( h + \frac{gx^2}{2u^2} \right) \implies 1 = \frac{2gh}{u^2} + \frac{g^2 x^2}{u^4} $$ $$ \frac{g^2 x^2}{u^4} = 1 – \frac{2gh}{u^2} = \frac{u^2 – 2gh}{u^2} $$ $$ x = \frac{u}{g} \sqrt{u^2 – 2gh} $$ Now substitute $x$ into the angle equation $\tan\theta = \frac{u^2}{gx}$: $$ \tan\theta = \frac{u^2}{g \left( \frac{u}{g} \sqrt{u^2 – 2gh} \right)} = \frac{u}{\sqrt{u^2 – 2gh}} $$ We need $\theta$. To convert $\tan\theta$ to $\sin\theta$, use the triangle geometry (Opposite $u$, Adjacent $\sqrt{u^2 – 2gh}$): $$ \text{Hypotenuse} = \sqrt{u^2 + (u^2 – 2gh)} = \sqrt{2u^2 – 2gh} $$ $$ \sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{u}{\sqrt{2(u^2 – gh)}} $$ $$ \theta = \sin^{-1} \left( \frac{u}{\sqrt{2(u^2 – gh)}} \right) $$
First, find $x_{max}$ from the discriminant equation: $$ x^2 = \frac{2gx^2}{u^2} \left( h + \frac{gx^2}{2u^2} \right) \implies 1 = \frac{2gh}{u^2} + \frac{g^2 x^2}{u^4} $$ $$ \frac{g^2 x^2}{u^4} = 1 – \frac{2gh}{u^2} = \frac{u^2 – 2gh}{u^2} $$ $$ x = \frac{u}{g} \sqrt{u^2 – 2gh} $$ Now substitute $x$ into the angle equation $\tan\theta = \frac{u^2}{gx}$: $$ \tan\theta = \frac{u^2}{g \left( \frac{u}{g} \sqrt{u^2 – 2gh} \right)} = \frac{u}{\sqrt{u^2 – 2gh}} $$ We need $\theta$. To convert $\tan\theta$ to $\sin\theta$, use the triangle geometry (Opposite $u$, Adjacent $\sqrt{u^2 – 2gh}$): $$ \text{Hypotenuse} = \sqrt{u^2 + (u^2 – 2gh)} = \sqrt{2u^2 – 2gh} $$ $$ \sin\theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{u}{\sqrt{2(u^2 – gh)}} $$ $$ \theta = \sin^{-1} \left( \frac{u}{\sqrt{2(u^2 – gh)}} \right) $$
Step 4: Solving for Case (b) – Target at depth $h$ ($y = -h$)
Similarly, substituting $y = -h$ into the discriminant yields: $$ x = \frac{u}{g} \sqrt{u^2 + 2gh} $$ The angle relation remains $\tan\theta = \frac{u^2}{gx}$: $$ \tan\theta = \frac{u}{\sqrt{u^2 + 2gh}} $$ Computing the hypotenuse: $\sqrt{u^2 + (u^2 + 2gh)} = \sqrt{2u^2 + 2gh}$. $$ \sin\theta = \frac{u}{\sqrt{2(u^2 + gh)}} $$ $$ \theta = \sin^{-1} \left( \frac{u}{\sqrt{2(u^2 + gh)}} \right) $$
Similarly, substituting $y = -h$ into the discriminant yields: $$ x = \frac{u}{g} \sqrt{u^2 + 2gh} $$ The angle relation remains $\tan\theta = \frac{u^2}{gx}$: $$ \tan\theta = \frac{u}{\sqrt{u^2 + 2gh}} $$ Computing the hypotenuse: $\sqrt{u^2 + (u^2 + 2gh)} = \sqrt{2u^2 + 2gh}$. $$ \sin\theta = \frac{u}{\sqrt{2(u^2 + gh)}} $$ $$ \theta = \sin^{-1} \left( \frac{u}{\sqrt{2(u^2 + gh)}} \right) $$
Final Answers:
(a) Landing at height $h$: $$ \theta = \sin^{-1} \left( \frac{u}{\sqrt{2(u^2 – gh)}} \right) $$ (b) Landing at depth $h$: $$ \theta = \sin^{-1} \left( \frac{u}{\sqrt{2(u^2 + gh)}} \right) $$
(a) Landing at height $h$: $$ \theta = \sin^{-1} \left( \frac{u}{\sqrt{2(u^2 – gh)}} \right) $$ (b) Landing at depth $h$: $$ \theta = \sin^{-1} \left( \frac{u}{\sqrt{2(u^2 + gh)}} \right) $$