Question

A boy playing on the roof of a $10\text{m}$ high building throws a ball with a speed of $10\text{m/s}$ at an angle of ${30}^o$ with the horizontal. How far from the throwing point will the ball be at the height of $10\text{m}$ from the ground?
$[g=10{\text{m/s}}^2,\sin {30}^0=\frac{1}{2},\cos {30}^0=\frac{\sqrt{3}}{2}]$

• A. $5.20\text{m}$
• B. $4.33\text{m}$
• C. $2.60\text{m}$
• D. $8.66\text{m}$

Answer 1

The correct option is D $8.66\text{m}$

$\text{The ball will be at the height of}10m\text{from the ground when it cover its maximum horizontal range.}$

Maximum horizontal range $R=\frac{u^2\sin 2\theta }{g}$$R=\frac{(10{)}^2\times 2\times \frac{\sqrt{3}}{2}\times \frac{1}{2}}{10}=8.66m$