Two bodies are projected with the same velocity. If one is projected at an angle of 30\(^{\circ}\) and the other at 45\(^{\circ}\) to the horizontal, then the ratio of maximum heights attained is
Show Hint
Remember that maximum height depends on the vertical component of the initial velocity squared. So, if angles are complementary, say $\theta$ and $90^\circ - \theta$, $\sin^2\theta$ will be different from $\sin^2(90^\circ-\theta) = \cos^2\theta$. This problem highlights the direct dependence on the square of the sine of the angle.
Step 1: Understanding the Question:
The problem asks for the ratio of maximum heights attained by two projectiles launched with the same initial velocity but at different angles of projection. Step 2: Key Formula or Approach:
The maximum height ($H$) for a projectile launched with initial velocity $u$ at an angle $\theta$ to the horizontal is given by:
\[ H = \frac{u^2 \sin^2\theta}{2g} \]
Since $u$ and $g$ are constant for both projections, $H \propto \sin^2\theta$. Step 3: Detailed Explanation:
Let $H_1$ be the maximum height for the first body projected at $\theta_1 = 30^{\circ}$, and $H_2$ for the second body projected at $\theta_2 = 45^{\circ}$.
The ratio of their maximum heights will be:
\[ \frac{H_1}{H_2} = \frac{\frac{u^2 \sin^2\theta_1}{2g}}{\frac{u^2 \sin^2\theta_2}{2g}} = \frac{\sin^2\theta_1}{\sin^2\theta_2} \]
Substitute the given angles:
\[ \frac{H_1}{H_2} = \frac{\sin^2(30^{\circ})}{\sin^2(45^{\circ})} \]
We know that $\sin(30^{\circ}) = \frac{1}{2}$ and $\sin(45^{\circ}) = \frac{1}{\sqrt{2}}$.
\[ \frac{H_1}{H_2} = \frac{(1/2)^2}{(1/\sqrt{2})^2} = \frac{1/4}{1/2} = \frac{1}{4} \times 2 = \frac{1}{2} \]
So, the ratio of maximum heights attained is 1:2. Step 4: Final Answer:
The ratio of maximum heights attained is 1:2.
