Question

A ball is thrown from the top of a tower with an initial velocity of $10\,ms^{-1}$ at an angle of $30^?$ with the horizontal. If it hits the ground at a distance of $17.3\,m$ from the base of the tower, the height of the tower is (Take $g= 10\,m s^{-2}$)

• A. $5\,m$
• B. $20\,m$
• C. $15\,m$
• D. $10\,m$

Answer 1

Answer: (D)

$10\,m$

Answer 2

Here, $\theta = 30^?$, $u = 10\,ms^{-1}$, $R = 17.3\,m$, $g= 10\,ms^{-2}$ For horizontal motion, $R = u\, cos \theta t$ or $t=\frac{R}{u\,cos\,\theta}$ $\frac{17.3}{10\,cos\,30^{?}}$ $\frac{17.3\times2}{10\times\sqrt{3}}$ $=\frac{17.3\times2}{10\times1.73}$ $=2\,s$ For vertical motion, $y=u\,sin\theta t-\frac{1}{2}gt^{2}$ $y=10\,sin\,30^{?}\times2-\frac{1}{2}\times10\times2^{2}$ $=10-20=-10\,m$. Height of tower $= 10\, m$.