Question

When a partical is projected from a tower of height $h$ with horizontal with velocity $u$ at an angle $\theta$ above the horizontal, then range of particle (on ground) is

• A. $[\frac{u\cos \theta }{g}+\sqrt{\frac{u^2{\sin }^2\theta }{g^2}+\frac{2h}{g}}]u\cos \theta$
• B. $[\frac{u\sin \theta }{g}+\sqrt{\frac{u^2{\sin }^2\theta }{g^2}+\frac{2h}{g}}]u\cos \theta$
• C. $\frac{u^2{\sin }^2\theta }{g^2}+\left[\sqrt{\frac{u^2{\sin }^2\theta }{g^2}+\frac{2h}{g}}\right]\cos \theta$
• D. $\frac{u^2{\sin }^2\theta }{g^2}$

Answer 1

The correct option is B $[\frac{u\sin \theta }{g}+\sqrt{\frac{u^2{\sin }^2\theta }{g^2}+\frac{2h}{g}}]u\cos \theta$

Apply kinematic equation of motion.

$v^2-u^2=2as$

$v=\sqrt{{(-u\sin \theta )}^2+2gh}$

Apply kinematic equation of motion

$V=U+at$

$t=\frac{V-U}{a}=\frac{1}{g}(\sqrt{{(-u\sin \theta )}^2+2gh}-(-u\sin \theta ))$

$t=(\sqrt{\frac{u^2{\sin }^2\theta }{g^2}+\frac{2h}{g}}+\frac{u\sin \theta }{g})$

Range, $R=u\cos \theta \times t=u\cos \theta (\sqrt{\frac{u^2{\sin }^2\theta }{g^2}+\frac{2h}{g}}+\frac{u\sin \theta }{g})$