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)

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 $= 10\, m$.