Question

A shell is fired vertically upwards with a velocity v┬╣ from the deck of a ship moving with a speed v┬▓.A person on the shore observes the motion of the shell as a parabola. It's horizontal range is given by:

Answer 1

in which class you study

Answer 2

Answer: $x\quad=\quad \frac { 2V_{ 1 }V_{ 2 } }{ g }$

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First let us consider the given diagram,  

Let us consider the horizontal range as x which $=\quad V_{ 2 }\quad \times \quad t$

Where,

't' is the time of motion  

                             $h\quad =\quad V_{ 1 }\quad \times \quad (t)\quad \longrightarrow \quad (1)$

We know that,

                             $h\quad =\quad \frac { 1 }{ 2 }\times g{ t }^{ 2 }\quad \longrightarrow (2)$

Comparing (1) and (2),

                             $V_{ 1 }\quad \times \quad t\quad =\quad \frac { 1 }{ 2 } \quad \times \quad g\quad \times \quad t^{ 2 }\\t\quad =\quad \frac { 2V_{ 1 } }{ g } \quad \longrightarrow (3)$

Hence, the horizontal range is equal to$x\quad =\quad \frac { V_{ 2 }\quad \times \quad (2V_{ 1 }) }{ g } \quad \\x\quad=\quad \frac { 2V_{ 1 }V_{ 2 } }{ g }$