Thank you for reporting, we will resolve it shortly
Q.
The range of a particle projected at an angle of 15° with the horizontal is 1.5 km. Its range when projected with the same velocity at an angle of 45° with the horizontal is
We know, $R = \frac{u^2 \, \sin \, 2\theta}{g}$
Here, $ \theta = 15^\circ, R = 1.5 \, km = 1500 \, m $
$ \therefore \:\:\: 1500 = \frac{u^2 }{g} \sin 30^\circ$ or , $ \frac{u^2}{g} = 3000 \, m$
Now for same velocity and $\theta = 45°$, range of the projectile would be,
$R' = \frac{u^2 \sin 90^\circ}{g} = \frac{u^2}{g} = 3000 \, m = 3 \, km $