The horizontal range of an oblique projectile is equal to the distance through which a projectile has to fall freely from rest to acquire a velocity equal to the velocity of projection in magnitude. The angle of projection is

Question

The horizontal range of an oblique projectile is equal to the distance through which a projectile has to fall freely from rest to acquire a velocity equal to the velocity of projection in magnitude. The angle of projection is (A) 75°  (B) 60°  (C) 45°  (D) 30°

Tardigrade Admin · Accepted Answer

Using v2-u2=2aS , we get S=(v2/2 g) Now, (v2 sin 2 θ /g)=(v2/2 g) or sin 2 θ = (1/2) Or sin 2θ =sinâ¡30° or θ =15° The other possible angle of projection is (90 ° - 15 ° ) , i.e., 75° .

Q. The horizontal range of an oblique projectile is equal to the distance through which a projectile has to fall freely from rest to acquire a velocity equal to the velocity of projection in magnitude. The angle of projection is

$7 5^{\circ}$

$6 0^{\circ}$

$4 5^{\circ}$

$3 0^{\circ}$

Using $v^{2} - u^{2} = 2 a S$ , we get
$S = \frac{v ^{2}}{2 g}$
Now, $\frac{v ^{2} s in 2 θ}{g} = \frac{v ^{2}}{2 g}$ or $s in 2 θ = \frac{1}{2}$
Or $s in 2 θ = s in 3 0^{\circ}$ or $θ = 1 5^{\circ}$
The other possible angle of projection is $(9 0^{\circ} - 1 5^{\circ})$ , i.e., $7 5^{\circ}$ .