Question

A projectile is launched from the ground, such that it hits a target on the ground which is 90 m away. The minimum velocity of the projectile to hit the target is (acceleration due to gravity $ g = 10\, \text{m/s}^2 $)

• A. \(10\, \text{m/s}\)
• B. \(16\, \text{m/s}\)
• C. \(60\, \text{m/s}\)
• D. \(30\, \text{m/s}\)

Hint:

Maximum range is achieved when angle of projection is \( 45^\circ \), and the range formula simplifies to \( R = \frac{u^2}{g} \).

Answer 1

The Correct Option is D

For a projectile to cover a horizontal range \( R \), the formula is: \[ R = \frac{u^2 \sin 2\theta}{g} \] The range will be maximum when \( \theta = 45^\circ \Rightarrow \sin 2\theta = 1 \). So, \[ 90 = \frac{u^2}{10} \Rightarrow u^2 = 900 \Rightarrow u = 30\, \text{m/s} \]