Question

Two particles are projected from the same point with the same speed $u$ such that they have the same range $R$, but different maximum heights, $h_1$ and $h_2$. Which of the following is correct ?

• A. $R^2 \, = \, 2 h_1 h_2$
• B. $R^2 \, = \, 16 h_1 h_2$
• C. $R^2 \, = \, 4 h_1 h_2$
• D. $R^2 \, = \, h_1 h_2$

Answer 1

The Correct Option is B

We need to find the relationship between the range \(R\) and the maximum heights \(h_1\) and \(h_2\) for two projectiles launched with the same speed. Here's the step-by-step reasoning:

1. The formula for the range \(R\) of a projectile launched with speed \(u\) and at an angle \(\theta\) is given by: \(R = \frac{u^2 \sin (2\theta)}{g}\), where \(g\) is the acceleration due to gravity.
2. The formula for the maximum height \(h\) is given by: \(h = \frac{u^2 \sin^2 \theta}{2g}\).
3. Since both particles have the same range \(R\) and same initial speed \(u\), they must be launched at complementary angles, \(\theta_1\) and \(90^\circ - \theta_1\), making \(h_1\) and \(h_2\) their respective maximum heights.
4. For complementary angles, the sum of the sine squares remains the same: \(\sin^2 \theta_1 + \sin^2 (90^\circ - \theta_1) = 1\).
5. Therefore, using the formula for maximum height: \(h_1 = \frac{u^2 \sin^2 \theta_1}{2g}\) and \(h_2 = \frac{u^2 \cos^2 \theta_1}{2g}\).
6. Multiplying the two equations for heights, we have: \(h_1 \cdot h_2 = \left(\frac{u^2 \sin^2 \theta_1}{2g}\right)\left(\frac{u^2 \cos^2 \theta_1}{2g}\right) = \frac{u^4 \sin^2 \theta_1 \cos^2 \theta_1}{4g^2}\).
7. The range equation when expressed in terms of trigonometric identities is: \(R = \frac{u^2 \sin 2\theta_1}{g} = \frac{u^2 \cdot 2\cdot \sin \theta_1 \cos \theta_1}{g}\).
8. Squaring both sides gives: \(R^2 = \left(\frac{u^2 \cdot 2 \sin \theta_1 \cos \theta_1}{g}\right)^2 = \frac{4u^4 \sin^2 \theta_1 \cos^2 \theta_1}{g^2}\).
9. Comparing both derived results, : \(R^2 = 16 h_1 h_2\)

Thus, the correct relationship between the range \(R^2\) and maximum heights \(h_1\) and \(h_2\) is given by \(R^2 = 16 h_1 h_2\).