A bullet fired at an angle of 30° with the horizontal hits the ground 3 km away. By adjusting its angle of projection, what is the farthest distance can one hope to hit.

Question

A bullet fired at an angle of 30° with the horizontal hits the ground 3 km away. By adjusting its angle of projection, what is the farthest distance can one hope to hit. (A) 3.46 km (B) 3.82 km (C) 4.12 km (D) 4.46 km

Tardigrade Admin · Accepted Answer

R=(u2 sin 2 θ/g) Here θ=30°, R=3 km , so (u2/g)=(3/ sin 60°) km But R max =(u2/g)=(3 × 2/√3)=2 √3=3.46 km

Q. A bullet fired at an angle of $3 0^{\circ}$ with the horizontal hits the ground $3 km$ away. By adjusting its angle of projection, what is the farthest distance can one hope to hit.

$3.46 km$

$3.82 km$

$4.12 km$

$4.46 km$

$R = \frac{u ^{2} s i n 2 θ}{g}$
Here $θ = 3 0^{\circ}, R = 3 km,$ so $\frac{u ^{2}}{g} = \frac{3}{s i n 6 0 ^{\circ}} km$
But $R_{m a x} = \frac{u ^{2}}{g} = \frac{3 \times 2}{3} = 23 = 3.46 km$

Q. A bullet fired at an angle of 30∘ with the horizontal hits the ground 3km away. By adjusting its angle of projection, what is the farthest distance can one hope to hit.

3.46km

3.82km

4.12km

4.46km