Question

If the height of a projectile at a time of 2 s from the beginning of motion is 60 m, then the time of flight of the projectile is (Acceleration due to gravity = 10 m/s\(^2\))

• A. 12 s
• B. 4 s
• C. 6 s
• D. 8 s

Hint:

For projectile motion, use \( h = u \sin\theta \, t - \frac{1}{2} g t^2 \) to find \( u \sin\theta \), then compute time of flight with \( T = \frac{2 u \sin\theta}{g} \).

Answer 1

The Correct Option is D

The height of a projectile at time \( t \) is: \[ h = u \sin\theta \, t - \frac{1}{2} g t^2 \] Given \( h = 60 \, \text{m} \) at \( t = 2 \, \text{s} \), \( g = 10 \, \text{m/s}^2 \): \[ 60 = u \sin\theta \cdot 2 - \frac{1}{2} \cdot 10 \cdot 2^2 \] \[ 60 = 2 u \sin\theta - \frac{1}{2} \cdot 10 \cdot 4 = 2 u \sin\theta - 20 \] \[ 2 u \sin\theta = 80 \] \[ u \sin\theta = 40 \] The time of flight is: \[ T = \frac{2 u \sin\theta}{g} \] \[ T = \frac{2 \cdot 40}{10} = 8 \, \text{s} \] Option (4) is correct. Options (1), (2), and (3) do not match.