Question

a particle is thrown with a initial velocity 2i+3j m/s .the horizontal range is​

Answer 1

THE RANGE OF PARTICLE IS 1.2 metres.

Given:

• Particle is thrown with a velocity of 2i + 3j m/s.

To find:

Horizontal range of the particle?

Calculation:

First of all, let's calculate the angle of projection.

$\tan( \theta) = \dfrac{ v_{y} }{ v_{x}}$

$\implies \tan( \theta) = \dfrac{ 3 }{ 2}$

So, the value of :

$\cos( \theta) = \dfrac{2}{ \sqrt{13} } \: and \: \sin( \theta) = \dfrac{3}{ \sqrt{13} }$

Now, let horizontal range be R :

$R = \dfrac{ {u}^{2} \sin(2 \theta) }{g}$

$\implies R = \dfrac{ {u}^{2} \times 2 \sin( \theta) \cos( \theta) }{g}$

$\implies R = \dfrac{ {( \sqrt{13}) }^{2} \times 2 \times \dfrac{2}{ \sqrt{13} } \times \dfrac{3}{ \sqrt{13} } }{10}$

$\implies R = \dfrac{ 13 \times 2 \times \dfrac{2}{ \sqrt{13} } \times \dfrac{3}{ \sqrt{13} } }{10}$

$\implies R = \dfrac{ 2 \times 2 \times 3 }{10}$

$\implies R = 1.2 \: metres$

The range of particle is 1.2 metres.