Question

A ball is thrown from a point in level with velocity $u$ and at a horizontal distance $r$ from the top of a tower of height $h$.
At what horizontal distance $x$ from the foot of the tower does the ball hit the ground?

• A. $\frac{u\cos \theta }{g}\left\{\right(u^2{\sin }^2\theta +gh{)}^{1/2}-u\sin \theta \}$
• B. $\frac{u\sin \theta }{g}\left\{\right(u^2{\cos }^2\theta +2gh{)}^{1/2}-u\cos \theta \}$
• C. $\frac{u\sin \theta }{g}\left\{\right(u^2{\cos }^2\theta +gh{)}^{1/2}-u\cos \theta \}$
• D. $\frac{u\sin \theta }{g}\left\{\right(u^2{\cos }^2\theta +gh{)}^{1/2}-u\cos \theta \}$

Answer 1

Answer: (A)

$\frac{u\cos \theta }{g}\left\{\right(u^2{\sin }^2\theta +gh{)}^{1/2}-u\sin \theta \}$

Here $r$ will become range
$r=\frac{u^2\sin 2\theta }{g}\Rightarrow gr=u^2\sin 2\theta$
$h=u\sin \theta t+\frac{1}{2}g{t}^2$ (Fig.)
$x=u\cos \theta t$
$t=\frac{x}{u\cos \theta }$
$h=u\sin \theta \frac{x}{u\cos \theta }+\frac{1}{2}g\frac{x^2}{u^2{\cos }^2\theta }$
$g{x}^2+2u^2\sin \theta \cos \theta x-2h{u}^2{\cos }^2\theta =0$
$x=\frac{u\cos \theta }{g}[\sqrt{u^2{\sin }^2\theta +2gh}-u\sin \theta ]$.