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.

Answer 2

Step 1: Recognize the Standard Projectile Trajectory Equation
The path of a projectile launched at an angle \( \theta \) with initial speed \( u \) is described by:
\[ y = x \tan \theta - \frac{g}{2 u^2 \cos^2 \theta} x^2. \] Comparing this to the given equation:
\[ y = 3x - 0.8 x^2, \] we can match terms:
\[ \tan \theta = 3, \quad \text{and} \quad \frac{g}{2 u^2 \cos^2 \theta} = 0.8. \]
Step 2: Calculate Initial Velocity \( u \)
First, express \( \sin \theta \) using \( \tan \theta \):
\[ \sin \theta = \frac{\tan \theta}{\sqrt{1 + \tan^2 \theta}} = \frac{3}{\sqrt{1 + 9}} = \frac{3}{\sqrt{10}}. \] Next, rearrange the relation involving \( u \):
\[ 0.8 = \frac{g}{2 u^2 \cos^2 \theta} \implies u^2 \cos^2 \theta = \frac{g}{2 \times 0.8} = \frac{10}{1.6} = 6.25. \] Since \( \cos^2 \theta = \frac{1}{1 + \tan^2 \theta} = \frac{1}{10} \), we find:
\[ u^2 = \frac{6.25}{0.1} = 62.5. \] Therefore, the initial speed is:
\[ u = \sqrt{62.5} \approx 7.9 \text{ m/s}. \]
Step 3: Find the Time of Flight \( T \)
Using the formula for total time of flight:
\[ T = \frac{2 u \sin \theta}{g} = \frac{2 \times 7.9 \times \frac{3}{\sqrt{10}}}{10}. \] Approximating \( \frac{3}{\sqrt{10}} \approx 0.95 \), we get:
\[ T \approx \frac{2 \times 7.9 \times 0.95}{10} = \frac{15.01}{10} = 1.5 \text{ s}. \]
Final Answer:
Thus, the time of flight is:
\[ \boxed{1.5 \text{ seconds}}. \]