Question

A package is dropped from a helicopter moving upward at 15 m/s. If it takes 16.0 s before the package strikes the ground, how high above the ground was the package when it was realeased?

Answer 1

Concepts and reason

The concepts used to solve the question is the projectile motion and the equation of the motion.

Initially use the equations of the motion along the vertical direction and then determine the height of the package from the ground.

Fundamentals

An object is said to be projectile motion when only one force is acting on it. The only force acting on the object during motion is the force of gravity. The force of gravity acts in a downward direction. The object moves with the acceleration equal to the acceleration due to gravity.

The acceleration with which the particle in projectile motion is moving is equal to the acceleration due to gravity.

Therefore,

${a_{\rm{y}}} = \pm g$

The sign of the acceleration is taken as positive when the motion of the particle is along the force of the gravity.

A body is said to be in motion when an external force is applied to it and therefore the body changes its position. The body’s motion can be explained in terms of the position, velocity, time and acceleration.

The equations of the motion are the set of equations that describe the position and the velocity of the body as a function of time.

The equations of motion are defined as,

$v = u + at$

Here, $v$ is the final velocity of the body, $u$ is the initial velocity of the body, $a$ is the acceleration of the body and $t$ is the time.

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

Here, $S$ is the displacement of the body.

${v^2} - {u^2} = 2aS$

The height of the package is,

$h - {h_0} = ut + \frac{{g{t^2}}}{2}$ …… (1)

Here, ${h_0}$ is the initial height and $g$ is the acceleration due to gravity.

Substitute $0$ for ${h_0}$ , $15{\rm{ m/s}}$ for $u$ , $16{\rm{ s}}$ for $t$ and $- 9.8{\rm{ m/}}{{\rm{s}}^2}$ for $g$ in the equation (1).

$\begin{array}{c}\\h - {h_0} = ut + \frac{{g{t^2}}}{2}\\\\h - 0 = \left( {15{\rm{ m/s}}} \right)\left( {16{\rm{ s}}} \right) + \frac{{\left( { - 9.8{\rm{ m/}}{{\rm{s}}^2}} \right){{\left( {16{\rm{ s}}} \right)}^2}}}{2}\\\\ = 240{\rm{ m}} - 1254.4{\rm{ m}}\\\\{\rm{ = }} - {\rm{1014}}{\rm{.4 m}}\\\end{array}$

Here, a negative sign indicates that the package falls down

Ans:

The height of the package from the ground is ${\bf{1014}}{\bf{.4 m}}$

Answer 2

The time to V = 0 derives from Vf = Vo - gt or 0 = 15 - 9.8t yielding t(up) = 1.53sec. resulting in t(dwn) = 16 - 1.53 = 14.47 sec.

h(dwn) then derives from h = Vo(t) + g(t^2)/2 or h = 0 + 4.9t(dwn)^2 yielding h(dwn) = 4.9(14.47)^2 = 1025.96 met.

h(up) then derives from h(up) = 15(1.53) - 4.9(14.47)^2 = 11.48 met.

The vertical rise from ejection to zero vertical velocity derives from h(up) = 15(1.53) - 4.9(1.53)^2 = 11.48 met.

Therefore, the altitude of release is h(rel) = 1025.96 - 11.48 = 1014.48 met.