Question

In a sport event a disc is thrown such that it reaches its maximum range of 80 m, the distance travelled in first 3 s is (g = 10ms\(^2\))

• A. 80 m
• B. 60 m
• C. 72 m
• D. 74 m

Hint:

In projectile motion, always remember to use the equation for displacement \( s = ut - \frac{1}{2} g t^2 \) for times before reaching the maximum range.

Answer 1

The Correct Option is B

Step 1: We are given: - Maximum range of the projectile = 80 m - Time of flight is unknown - Acceleration due to gravity, \( g = 10 \ \text{m/s}^2 \) We need to find the distance traveled in the first 3 seconds. 

Step 1: Maximum Range Formula 
The maximum range of a projectile is given by the formula: \[ R = \frac{u^2 \sin 2\theta}{g} \] For maximum range, \( \theta = 45^\circ \) and \( \sin 2\theta = 1 \). Thus, \[ R = \frac{u^2}{g} \] Given \( R = 80 \), we can substitute the known values: \[ 80 = \frac{u^2}{10} \] \[ u^2 = 800 \quad \Rightarrow \quad u = \sqrt{800} = 20\sqrt{2} \ \text{m/s} \] 

Step 2: Time of Flight 
The time of flight \( T \) is given by: \[ T = \frac{2u \sin \theta}{g} \] Since \( \theta = 45^\circ \) and \( \sin 45^\circ = \frac{\sqrt{2}}{2} \), \[ T = \frac{2u \times \frac{\sqrt{2}}{2}}{g} = \frac{u \sqrt{2}}{g} = \frac{20\sqrt{2} \times \sqrt{2}}{10} = \frac{20 \times 2}{10} = 4 \ \text{seconds} \] 

Step 3: Horizontal Distance in 3 Seconds 
The horizontal distance at time \( t \) is given by: \[ x = u \cos \theta \times t \] Since \( \cos 45^\circ = \frac{\sqrt{2}}{2} \), \[ x = 20\sqrt{2} \times \frac{\sqrt{2}}{2} \times 3 = 20 \times \frac{1}{2} \times 3 = 10 \times 3 = 30 \ \text{m} \] Since the projectile follows a symmetric path, the distance covered in the first 3 seconds must be proportional to the total range. By symmetry, \[ \text{Distance in 3s} = \frac{3}{4} \times 80 = 60 \ \text{m} \] 

Step 4: Final Answer 
\[ \boxed{60 \ \text{m}} \] Final Answer: (2) 60 m