Question

a helicopter is ascending vertically with a speed of 5.10 m/s. At a height of 105 m above the earth, a package is dropped from a window. How much time does it take for the package to reach the ground?(hint v0 for the package equals the speed of the helicopeter.

Answer 1

Concepts and reason

The concepts used to solve this problem are kinematic equations.

In classical physics, the position, velocity and acceleration of any object is calculated by using the Kinematic equations. Calculate the time taken by the package to reach the ground by solving the position time equation.

Fundamentals

If an object moves, it must follow the equations of motion. The expressions of kinematic equations are,

$\begin{array}{l}\\v = u + at\\\\S = ut + \frac{1}{2}a{t^2}\\\\{v^2} = {u^2} + 2aS\\\end{array}$

Here, $v$ is the final velocity, $u$ is the initial velocity, $a$ is the acceleration , $S$ is the distance covered by the object and $t$ is the time taken.

The standard form of a quadratic equation is,

$a{x^2} + bx + c = 0$

The solution of equation is,

$x = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Here, $a,b{\rm{ and }}c$ are constants.

Calculate the time taken by the package to reach the ground.

The expression for position time equation is,

$S = ut + \frac{1}{2}a{t^2}$

Here, $u$ is the initial velocity, $a$ is the acceleration, $S$ is the distance covered by the object and $t$ is the time taken.

Rearrange it for vertical direction as taking positive signs for downwards direction,

$H = ut + \frac{1}{2}g{t^2}$

Here, $H$ is the height and $g$ is acceleration due to gravity.

Substitute $105{\rm{ m}}$ for $H$ , $- 5.10{\rm{ m/s}}$ for $u$ and $9.8{\rm{ m/}}{{\rm{s}}^2}$ for $g$ .

$105{\rm{ m}} = \left( { - 5.10{\rm{ m/s}}} \right)t + \frac{1}{2}\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right){t^2}$

Rearrange it as follows,

$\left( {4.9{\rm{ m/}}{{\rm{s}}^2}} \right){t^2} - \left( {5.10{\rm{ m/s}}} \right)t - 105{\rm{ m}} = 0$

It is quadratic equation. The solution of this equation is,

$t = \frac{{ - \left( { - 5.10{\rm{ m/s}}} \right) \pm \sqrt {{{\left( { - 5.10{\rm{ m/s}}} \right)}^2} - 4\left( {4.9{\rm{ m/}}{{\rm{s}}^2}} \right)\left( { - 105{\rm{ m}}} \right)} }}{{2\left( {4.9{\rm{ m/}}{{\rm{s}}^2}} \right)}}$

It gives,

$t = 5.18{\rm{ s or }} - {\rm{4}}{\rm{.13 s}}$

The time cannot be negative. So, ignore the negative value of time.

Therefore, the time taken by the package to reach the ground is $5.18{\rm{ s}}$ .

Ans:

The time taken by the package to reach the ground is $5.18{\rm{ s}}$ .