Question

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

Answer 1

$\Large{\underline{\underline{\bf{Solution :}}}}$

We know that,

$\Large{\implies{\boxed{\boxed{\rm{R = \frac{u^2 sin2\theta}{g}}}}}}$

Where,

• g = 10 $\tt{ms^{-2}}$
• u = 10 $\tt{ms^{-1}}$
• Theta($\theta)$ = 30°

______________[Put Values]

$\tt{→R = \frac{(10)^2 \times Sin2(30^{\circ})}{10}} \\ \\ \tt{→R = \frac{\cancel{100} \times Sin60^{\circ}}{\cancel{10}}} \\ \\ \tt{→R = \cancel{10} \times \frac{\sqrt{3}}{\cancel{2}}} \\ \\ \tt{→R = 5 \sqrt{3} \: m} \\ \\ \Large{\implies{\boxed{\boxed{\rm{R = 5\sqrt{3} \: m}}}}}$