Question

A ball is projected with velocity 10 m/s at an angle of 30 degree with the horizontal surface. The range of the projectile is​

Answer 1

Answer is at the above image.

Answer 2

Given:

A ball is projected with velocity 10 m/s at an angle of 30 degree with the horizontal surface.

To find:

Range of the Projectile

Calculation:

This is an example of Ground-Ground Projectile.

General expression for range of a Projectile thrown at an angle of $\theta$ and initial velocity "u" is as follows:

$\boxed{ \bold{ \therefore \: range = \dfrac{ {u}^{2} \sin(2 \theta) }{g} }}$

$= > \: range = \dfrac{ {(10)}^{2} \sin(2 \times {30}^{ \circ} ) }{g}$

$= > \: range = \dfrac{ {(10)}^{2} \sin({60}^{ \circ} ) }{g}$

$= > \: range = \dfrac{100 \sin({60}^{ \circ} ) }{g}$

$= > \: range = \dfrac{100 \sin({60}^{ \circ} ) }{10}$

$= > \: range = 10 \sin({60}^{ \circ} )$

$= > \: range = 10 \times \dfrac{ \sqrt{3} }{2}$

$= > \: range = 5 \sqrt{3} \: metres$

So, range of Projectile is 5√3 m/s.