Question

The equation of projectile motion is given by \( y = 3x - 0.8x^2 \). The time of flight of the projectile is (Acceleration due to gravity \( g = 10 \) m/s²):

• A. \( 1.5 \) s
• B. \( 3 \) s
• C. \( 2 \) s
• D. \( 2.5 \) s

Hint:

For projectile motion problems, comparing with the standard trajectory equation allows us to extract key parameters such as \( u \) and \( \theta \), leading to an accurate determination of flight time.

Answer 1

The Correct Option is A

Step 1: Identify Standard Equation of Trajectory The trajectory of a projectile is generally given by: \[ y = x \tan \theta - \frac{g}{2u^2 \cos^2 \theta} x^2. \] Comparing this with the given equation: \[ y = 3x - 0.8x^2, \] we identify: \[ \tan \theta = 3, \quad \frac{g}{2u^2 \cos^2 \theta} = 0.8. \]
Step 2: Determine Initial Velocity \( u \) The time of flight \( T \) for projectile motion is given by: \[ T = \frac{2u \sin \theta}{g}. \] We first find \( \sin \theta \): \[ \sin \theta = \frac{\tan \theta}{\sqrt{1 + \tan^2 \theta}} = \frac{3}{\sqrt{1 + 9}} = \frac{3}{\sqrt{10}}. \] Now, solving for \( u^2 \): \[ 0.8 = \frac{10}{2u^2 \cos^2 \theta} \Rightarrow u^2 \cos^2 \theta = \frac{10}{2 \times 0.8} = \frac{10}{1.6} = 6.25. \] Since \( \cos^2 \theta = \frac{1}{10} \), \[ u^2 = \frac{6.25}{0.1} = 62.5. \] Thus, \[ u = \sqrt{62.5} \approx 7.9 \text{ m/s}. \]
Step 3: Compute Time of Flight \[ T = \frac{2u \sin \theta}{g} = \frac{2 \times 7.9 \times \frac{3}{\sqrt{10}}}{10}. \] Approximating, \[ T = \frac{2 \times 7.9 \times 0.95}{10} \approx 1.5 \text{ s}. \] % Final Answer Thus, the correct answer is option (1): \( 1.5 \) s.