CBSE-XI-Physics
03: Rest and Motion: Kinematics
- #10The 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
-(a) 25 m
(b) 37 m
(c) 50 m
(d) 100 mdigAnsr: dAns : (d) 100 m
For the same u range, `` R\propto \,\mathrm{\,sin \,}\left(2\theta \right)``.
So,
`` \frac{{R}_{1}}{{R}_{2}}=\frac{\,\mathrm{\,sin \,}\left(2{\theta }_{1}\right)}{\,\mathrm{\,sin \,}\left(2{\theta }_{2}\right)}``
`` \Rightarrow {R}_{2}=50\times \frac{\,\mathrm{\,sin \,}\left(90\right)}{\,\mathrm{\,sin \,}\left(30\right)}=100\,\mathrm{\,m \,}``