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Q.
The horizontal range of a projectile fired at an angle of $15^°$ is $50\,m$. If it is fired with the same speed at an angle of $45^°$, its range will be
Motion in a Plane
Solution:
Horizontal range,
$ R=\frac{u^{2}\,sin\,2\theta}{g}$
For the same speed,
$R\,\propto\,sin\,2\theta$
$\therefore \frac{R_{1}}{R_{2}}=\frac{sin\,2\times15^{°}}{sin\,2\times45^{°}}$
$=\frac{sin\,30^{°}}{sin\,90^{°}}$
or $R_{2}=R_{1} \frac{sin\,90^{°}}{sin\,30^{°}}$
$=50\,m\times\frac{1}{\left(\frac{1}{2}\right)}$
$=100\,m$