The range of projectile is 1 .5 km when it is projected at an angle of 15° with horizontal. What will be its range when it is projected at an angle

The maximum horizontal distance travelled by the projectile is called the horizontal range (R) of the projectile.

Correct Option is (B) \(45^\circ\) Range of projectile , \(R = \frac{u^2 \ sin2 \ \theta}{g}\) \(R_{max} = \frac{u^2 \ sin2 \ \theta}{g}\) \(\because \theta = 45^\circ\) \(R_{max} = \frac{u^2 \ sin 90^\circ}{g}\) \(R_{max} = \frac{u^2}{g}\)

Correct option is: (D) greater than 45°

Correct Option is (C) \(60^\circ\) Given, H = 0.433R We know that , \(H = \frac{u^2sin^2 \theta}{2g} , R = \frac{u^2sin^2 \theta}{g}\) \( \frac{u^2sin^2 \theta}{g} = 0.433 \frac{u^2sin^2 \theta}{g}\) \(\frac{sin^2 \theta}{2g} = 0.433 . 2 \ sin...

Correct Option is (B) 125m \(H_{max} = \frac{u^2}{2g}\) \(H_{max} = \frac{50 \times 50}{2 \times 10}\) \(H_{max} = 125 m\)

Correct option is: (B) same angle

Correct Answer - (i)`theta=45^(@)` (ii)`R=(u^(2))/(g),H=(u^(2))/(4g)` (iii) 1 : 1