Question

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

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

Answer 1

Answer: (D)

Here, $\theta =30^{°}$,
$u=10m{s}^{-1}$,
$R=17.3m$,
$g=10m{s}^{-2}$
For horizontal motion,
$R=ucos\theta t$
or $t=\frac{R}{ucos\theta }$
$\frac{17.3}{10cos30^{°}}$
$\frac{17.3\times 2}{10\times \sqrt{3}}$
$=\frac{17.3\times 2}{10\times 1.73}$
$=2s$
For vertical motion,
$y=usin\theta t-\frac{1}{2}g{t}^2$
$y=10sin30^{°}\times 2-\frac{1}{2}\times 10\times 2^2$
$=10-20=-10m$.
Height of tower $=10m$.