Question

Solve it.................​

Answer 1

$\huge\bf{\red{\overbrace{\underbrace{\purple{Given:}}}}}$

★The speed of the helicopter is 'u'.

★It is flying at a height 'h'.

$\huge\bf{\red{\overbrace{\underbrace{\purple{To\:\:Find:}}}}}$

★The distance at which the packet should be dropped so that it reaches to the victim.

$\huge\bf{\red {\overbrace{\underbrace{\purple{Answer:}}}}}$

$\huge{\red{\sf{Given}}}$$\begin{cases}\longrightarrow Initial velocity = u\\ \longrightarrow Speed = u \end{cases}$

We know,

$\large\red{\boxed{\purple{\bf{Distance=Speed\times time }}}}$

Let. the required distance be d.

$\implies d =u\times t$

${\underline{\boxed{\red{.\degree. t =\dfrac{d}{u}}}}}$

______________________________________

Now, In vertical direction,

The vertical velocity is 0 .

So,

$\large\red{\boxed{\purple{s=ut+\frac{1}{2}at^{2}}}}$

$\implies s = 0\times t +\dfrac{1}{2}gt^{2}$

$\implies h =\dfrac{gt^{2}}{2}$

$\implies h =\dfrac{g\times (d) ^{2}}{2\times (u) ^{2}}$

$\implies h =\dfrac{g\times d^{2}}{2\times u ^{2}}$

$\implies d^{2}=\dfrac{2h \times u^{2}}{g}$

$\implies d =\sqrt{\dfrac{2h \times u^{2}}{g}}$

${\underline{\boxed{\purple{.\degree. d=u\sqrt{\dfrac{2h}{g}}}}}}$