Q.Two particles are projected in air with speed $v_o$ at angles $\theta_1$ and $\theta_2$ (both acute) to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then tick the right choices (Note: more than one of the given options may be correct.)

Concept: Gravitational Potential Energy — the maximum height depends only on the vertical component of initial velocity.

Reasoning:

1. Maximum height: $H = \frac{v_o^2 \sin^2\theta}{2g}$.

   Given $H_1 > H_2$, we have $\sin^2\theta_1 > \sin^2\theta_2 \Rightarrow \sin\theta_1 > \sin\theta_2$.

   Since both angles are acute, $\theta_1 > \theta_2$. So (A) is correct.

2. Time of flight: $T = \frac{2v_o \sin\theta}{g}$.

   With $\theta_1 > \theta_2$, $T_1 > T_2$. So (B) is correct.

3. Horizontal range: $R = \frac{v_o^2 \sin 2\theta}{g}$. …

Let’s start with the core idea. In projectile motion, the maximum height reached is determined solely by the vertical component of the initial velocity. The horizontal component does nothing to lift the particle — it only carries it sideways. So when we are told that the first particle goes higher than the second, we immediately know something about their vertical launches.

The vertical component for a particle projected at speed $v_o$ and angle $\theta$ is $v_o \sin \theta$. The maximum height $H$ is given by:

Since $v_o$ and $g$ are the same for both, $H_1 > H_2$ means $\sin^2 \theta_1 > \sin^2 \theta_2$. For acute angles ($0 < \theta < 90^\circ$), $\sin \theta$ increases with $\theta$, so this implies $\theta_1 > \theta_2$. That gives us option (A) as correct.

Now let’s go through each option step by step.

1. Angle of projection: $\theta_1 > \theta_2$

   As argued above, height depends on $\sin^2 \theta$. Since $\sin \theta$ is increasing for acute angles, a greater height forces a greater angle. So (A) is correct.

2. Time of flight: $T_1 > T_2$

   Time of flight is $T = \frac{2 v_o \sin \theta}{g}$. This also depends only on the vertical component. Since $\theta_1 > \theta_2$, $\sin \theta_1 > \sin \theta_2$, so $T_1 > T_2$. Option (B) is correct.

3. Horizontal range: $R_1 > R_2$

   Range is $R = \frac{v_o^2 \sin 2\theta}{g}$. This depends on $\sin 2\theta$, which is not monotonic over acute angles — it increases up to $45^\circ$ and then decreases. So even though $\theta_1 > \theta_2$, we cannot say which has the larger $\sin 2\theta$ without knowing the actual values. For example, if $\theta_1 = 60^\circ$ and $\theta_2 = 30^\circ$, then $\sin 120^\circ = \sin 60^\circ$, so ranges are equal. If $\theta_1 = 50^\circ$ and $\theta_2 = 40^\circ$, then $\theta_1$ gives a smaller range. So (C) is not necessarily true. …

Concept: Monotonic Functions Can Be Compared by Their Inputs Alone

Method: Calculus of Monotonicity (Sign of the Derivative), Applied Once to Each of $H(\theta)$, $T(\theta)$, $R(\theta)$

Rather than plugging specific numbers into $H$, $T$, $R$ and comparing the resulting values case by case, this method checks whether each quantity is a strictly increasing function of $\theta$ on the acute range $(0^\circ,90^\circ)$. If a function is strictly increasing, a bigger output necessarily means a bigger input -- so the given fact "$H_1>H_2$" can be translated straight into "$\theta_1>\theta_2$" (and then propagated to $T$) without any numerical substitution at all.

Step 1 -- Is $H(\theta)$ increasing on $(0^\circ,90^\circ)$?

For $\theta\in(0^\circ,90^\circ)$, $2\theta\in(0^\circ,180^\circ)$, where $\sin2\theta>0$ throughout. So $dH/d\theta>0$ everywhere: $H(\theta)$ is strictly increasing.

Step 2 -- Use monotonicity, not substitution, to get (A)

Because $H$ is strictly increasing and one-to-one on $(0^\circ,90^\circ)$, it has an inverse -- meaning $H(\theta_1)>H(\theta_2) \iff \theta_1>\theta_2$, with no other possibility. Given $H_1>H_2$, this immediately forces $\theta_1>\theta_2$, purely from the shape of the function, without ever writing a $\sin$ value. (A) is correct.

Step 3 -- Is $T(\theta)$ increasing on $(0^\circ,90^\circ)$?

For $\theta\in(0^\circ,90^\circ)$, $\cos\theta>0$, so $dT/d\theta>0$ throughout: $T(\theta)$ is also strictly increasing.

Step 4 -- Chain the two monotonic results together to get (B)

From Step 2, $\theta_1>\theta_2$. Since $T$ is strictly increasing (Step 3), this input inequality propagates straight through: $\theta_1>\theta_2 \Rightarrow T(\theta_1)>T(\theta_2)$, i.e. $T_1>T_2$. (B) is correct -- reached purely by chaining two "increasing function" facts, never touching an explicit numeric time.

Step 5 -- Is $R(\theta)$ increasing on $(0^\circ,90^\circ)$? (This is where the chain breaks)

$R(\theta) = \frac{v_0^2\sin2\theta}{g}, \qquad \frac{dR}{d\theta} = \frac{2v_0^2\cos2\theta}{g}$ …