Question

On a distant planet whose gravitational acceleration is 10 ms⁻² a point mass is projected with a velocity of 1 ms⁻¹ with an angle 0.25π rad with the horizontal. The length 'L' of parabolic arc traced by the projectile is computed as

• A. L = 10⁻¹($\sqrt{2}^{-1}$ + 2⁻¹ ln(1 + $\sqrt{2}$)) m
• B. L = 10⁻¹($\sqrt{2}^{-1}$ + 2⁻¹ ln(1 + $\sqrt{5}$)) m
• C. L = 10⁻¹($\sqrt{2}$ + 2⁻¹ ln(1 + $\sqrt{3}$)) m
• D. L = 10⁻¹($\sqrt{2}$ + 2⁻¹ ln(1 + $\sqrt{2}$)) m

Answer 1

Answer: (A)

L = 10⁻¹($\sqrt{2}^{-1}$ + 2⁻¹ ln(1 + $\sqrt{2}$)) m

Answer 2

The length of the parabolic arc traced by the projectile is calculated using the trajectory equation and arc length formula, resulting in $L = 10^{-1}(\sqrt{2}^{-1} + 2^{-1} \ln(1 + \sqrt{2})) \, \text{m}$.

The detailed solution involves:

1. Trajectory Equation: $y = x \tan\theta - \frac{gx^2}{2v_0^2 \cos^2\theta}$
2. Given Values: $v_0 = 1 \, \text{ms}^{-1}$, $g = 10 \, \text{ms}^{-2}$, $\theta = \frac{\pi}{4}$
3. Simplified Trajectory: $y = x - 10x^2$
4. Range: $R = \frac{1}{10}$
5. Arc Length Formula: $L = \int_0^R \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$
6. Derivative: $\frac{dy}{dx} = 1 - 20x$
7. Integral: $L = \int_0^{1/10} \sqrt{1 + (1 - 20x)^2} dx$
8. Substitution: $u = 1 - 20x$, $du = -20 dx$
9. Transformed Integral: $L = \frac{1}{20} \int_{-1}^1 \sqrt{1 + u^2} du$
10. Standard Integral: $\int \sqrt{1 + u^2} du = \frac{u}{2}\sqrt{1+u^2} + \frac{1}{2}\ln|u + \sqrt{1+u^2}| + C$
11. Definite Integral Evaluation: $\int_{-1}^1 \sqrt{1 + u^2} du = \sqrt{2} + \ln(\sqrt{2} + 1)$
12. Final Result: $L = \frac{1}{20} \left(\sqrt{2} + \ln(\sqrt{2} + 1)\right) = 10^{-1} \left(\frac{1}{\sqrt{2}} + \frac{1}{2}\ln(1 + \sqrt{2})\right) \, \text{m}$