Question

The velocity of projection of an oblique projectile is (6i+8j)m/s . The horizontal range of projectile is (g=10m/s)2

Answer 1

Since velocity is a vector quantity,

                                $v\Rightarrow =vxi=vyj$

Since it is a projectile, it has both x vector and y vector so the actual velocity becomes.

                                $v=\sqrt { \left( { x }^{ 2 }+{ y }^{ 2 } \right)  } \\=\sqrt { { 6 }^{ 2 }+{ 8 }^{ 2 } } \\=\sqrt { 100 } \\=10$

Now since it is a triangle we can find,

                                $sin\theta \quad and\quad cos\theta \Rightarrow sin\theta=\frac { 8 }{ 10 } \quad and \quad cos\theta=\frac { 6 }{ 10 }$

We know formula for range,

                                $R=\frac { ({ v }^{ 2 }sin2\theta) }{ g }$

                                $G=10\times \frac { m }{ { s }^{ 2 } }$

Given so,

                                $R=\frac { (2\times { v }^{ 2 }sin\theta cos\theta) }{ g }$

                                $=\frac { (2\times 10\times 10\times \frac { 8 }{ 10 } \times \frac { 6 }{ 10 } ) }{ 10 }$

                                $=\frac { 96 }{ 10 } =9.6m$