Question

A particle is released from a certain height of $600\text{m}$. Due to the wind, the particle gathers a horizontal velocity component of $ay$ where $a$ is $\sqrt{10}{\text{s}}^{-1}$ and $y$ is the vertical displacement of the particle from the point of release. What is the horizontal drift of the particle when it strikes the ground?
[Take $g=10{\text{m/s}}^2$]

• A. $7315\text{m}$
• B. $7215\text{m}$
• C. $7115\text{m}$
• D. $7015\text{m}$

Answer 1

The correct option is D $7015\text{m}$

Given: Height, $H=600\text{m}$
Horizontal velocity gained due to wind,
$v_x=ay=\sqrt{10}y$ at distance $y$ below point of release
The time of flight of the particle,
$t=\sqrt{\frac{2H}{g}}=\sqrt{\frac{2\times 600}{10}}\approx 11\text{s}$
Now, $v_x=\sqrt{10}y$
$\Rightarrow \frac{dx}{dt}=\sqrt{10}y$
$\Rightarrow {\int }_0^{x}dx=\sqrt{10}{\int }_0^{t}ydt$
$\Rightarrow {\int }_0^{x}dx=\sqrt{10}{\int }_0^t\frac{1}{2}g{t}^{2}dt$
$\Rightarrow {\int }_0^{x}dx=5\sqrt{10}{\int }_0^{11}{t}^{2}dt$
$\Rightarrow x=5\sqrt{10}{\left[\frac{t^3}{3}\right]}_0^{11}$
$\Rightarrow x=5\sqrt{10}[\frac{(11{)}^3}{3}-0]$
$\Rightarrow x\approx 7015\text{m}$
Hence, the option $\left(d\right)$ is correct.