Animated Solution for Physics - Kinematics: The ranges and heights for two projectiles projected with the same initial velocity at angles $42^{\circ }$ and $48^{\circ }$ with the horizontal are $R_1,R_2$ and $H_1,H_2$, respectively. Choose the correct option.

The problem of comparing the trajectories of two projectiles launched at different angles is a classic in kinematics. It beautifully illustrates the symmetry and mathematical elegance hidden within the equations of motion. Let's embark on this journey to decode the relationship between their ranges and maximum heights.

Imagine you are standing on a flat ground, holding two identical stones. You throw both of them with the exact same initial speed, $u$. However, you launch the first stone at an angle of $42^{\circ }$ and the second one at a slightly steeper angle of $48^{\circ }$.

Our goal is to figure out which stone travels further horizontally (the range, $R$) and which one reaches a greater vertical height (the maximum height, $H$).

Let's first tackle the horizontal range. The formula that governs the range of a projectile is given by:

Here, $u$ is the initial velocity, $\theta$ is the angle of projection, and $g$ is the acceleration due to gravity. Let's plug in our angles to see what happens.

For the first projectile (${\theta }_1=42^{\circ }$):

For the second projectile (${\theta }_2=48^{\circ }$):

At first glance, $\sin (84^{\circ })$ and $\sin (96^{\circ })$ might look different. But let's recall a beautiful property from trigonometry: $\sin (180^{\circ }-x)=\sin (x)$.

If we apply this to $96^{\circ }$, we get $\sin (96^{\circ })=\sin (180^{\circ }-84^{\circ })=\sin (84^{\circ })$.

This means that $R_1$ is exactly equal to $R_2$.

There is a powerful shortcut here: whenever two angles of projection add up to $90^{\circ }$ (they are complementary), and the initial speeds are the same, their horizontal ranges will always be identical. Since $42^{\circ }+48^{\circ }=90^{\circ }$, we could have instantly concluded that $R_1=R_2$.

Maximum Height

Now, let's shift our focus to the vertical motion. The formula for the maximum height attained by a projectile is:

Unlike the range formula which uses $\sin (2\theta )$, the height formula depends on ${\sin }^2\theta$. Let's set up the equations for our two projectiles:

To compare $H_1$ and $H_2$, we just need to compare $\sin (42^{\circ })$ and $\sin (48^{\circ })$.

In the first quadrant (from $0^{\circ }$ to $90^{\circ }$), the sine function is strictly increasing. This means that a larger angle will always have a larger sine value. Since $48^{\circ }>42^{\circ }$, it is guaranteed that $\sin (48^{\circ })>\sin (42^{\circ })$.

Consequently, the square of the sine will also be larger, leading us to the undeniable conclusion that $H_2>H_1$.

By breaking down the problem into horizontal and vertical components, we've uncovered the physical reality of the situation. The two projectiles will land at the exact same spot, but the one thrown at the steeper angle ($48^{\circ }$) will take a higher, more lofted path to get there.

Therefore, the correct relationship is $R_1=R_2$ and $H_1<H_2$.