Question

A projectile is fired with initial velocity u=(2i^+3j^)m/sec.Where i^ and j^ are unit vectors along horizontal and vertical direction respectively. The horizontal range of Projectile is?

Answer 1

Given:

A projectile is fired with initial velocity u=(2i^+3j^)m/sec.Where i^ and j^ are unit vectors along horizontal and vertical direction respectively.

To find:

Horizontal range of the projectile ?

Calculation:

The general expression for horizontal range of a projectile is :

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

Now, we can simplify as :

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

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

$\sf \implies \: R = \dfrac{2 \times \{u_{y} \} \times \{ u_{x}\} }{g}$

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

$\sf \implies \: R = \dfrac{12}{10}$

$\sf \implies \: R = 1.2 \: metres$

So, range of projectile is 1.2 metres.