# If a stone is dropped at an altitude of 174.9 m from a helicopter which is ascending with a velocity of 20.68 m/s, how long does the stone take to reach the ground?

## The problem I am having is that I am confused about using positive $g$ because gravity is directed towards the Earth and the stone is falling in that direction or if I should be using negative $g$ because the stone is falling opposite to the direction in which the helicopter is ascending, upwards?? It would really help if you guys could explain why g should be negative or positive. Thanks guys. :))

Aug 17, 2016

8.45 seconds.

#### Explanation:

The direction of 'g' when talking about acceleration depends on the coordinate system we define. For example if you were to define downwards as the positive 'y' then g would be positive. Convention is to take upwards as positive so g will be negative. This is what we shall use, also we take the ground as $y = 0$

$\textcolor{red}{\text{EDIT:}}$ I have added an approach using the kinematic equations you learn early on at the bottom. All I have done here is derive these using calculus but I appreciate you may not have covered it. Scroll down to the red title for the non calculus approach.

We can look at this much more closely by starting from scratch with Newton's second law. When the stone is dropped it has an initial velocity but the only force acting on it is due to gravity. We have defined upwards as the positive y direction so by Newton's second law we can write

$m \frac{{d}^{2} y}{{\mathrm{dt}}^{2}} = - m g$

$\frac{{d}^{2} y}{{\mathrm{dt}}^{2}} = - g$

This is because the stone will accelerate towards the earth, which we have defined as the negative direction.

Integrating this expression gives:

$\frac{\mathrm{dy}}{\mathrm{dt}} = - g t + C$

$\frac{\mathrm{dy}}{\mathrm{dt}} = y ' \left(t\right)$ is the velocity of the stone, so when we apply the initial velocity at $y ' \left(0\right) = + 20.68$ we arrive at

$20.68 = g \cdot 0 + C$

$\implies C = 20.68$

$\frac{\mathrm{dy}}{\mathrm{dt}} = 20.68 - g t$

This models the velocity and makes sense if you think about it. When it's released, it will have the same velocity as the helicopter and will thus move upwards for a time but as time progresses it will stop and then begin to fall.

To find displacement, we integrate again:

$y \left(t\right) = 20.68 t - \frac{1}{2} g {t}^{2} + C$

Apply initial condition $y \left(0\right) = 174.9$

$174.9 = 20.68 \cdot 0 - \frac{1}{2} g \cdot {0}^{2} + C$

$\implies C = 174.9$

$\therefore y \left(t\right) = 20.68 t - \frac{1}{2} g {t}^{2} + 174.9$

To solve for the time to reach the ground, set $y = 0$ and solve the quadratic:

$\frac{1}{2} g {t}^{2} - 20.68 t - 174.9 = 0$

This is definitely a job for the quadratic formula:

$t = \frac{20.68 \pm \sqrt{{20.68}^{2} - 4 \left(\frac{1}{2} g\right) \left(- 174.9\right)}}{g}$

Taking $g = 9.8 m {s}^{- 2}$

$t = 8.45 \mathmr{and} - 4.23$

We discard the negative solution so therefore the stone takes 8.45 seconds to hit the ground.

$\textcolor{red}{\text{No Calculus Approach}}$

We know that $v = {v}_{0} + a t$ where $v$ is the final velocity, ${v}_{0}$ is the initial velocity, $a$ is acceleration and $t$ is the time it's applied for.

As I said earlier, with an upwards coordinate system $g$ will be negative but the stone will initially move upwards because of it's initial velocity. We want to find the point at which it stops moving upwards:

Set $v = 0$

$0 = {v}_{0} - g t$

$\therefore t = {v}_{0} / g = \frac{20.68}{9.8}$

Now use

$S = {v}_{0} t + \frac{1}{2} a {t}^{2}$ again with $a = - g$

so $S = {v}_{0} \left({v}_{0} / g\right) - \frac{1}{2} g {\left({v}_{0} / g\right)}^{2}$

$S = {\left({v}_{0}\right)}^{2} / g - {v}_{0}^{2} / \left(2 g\right)$

$S = {\left(20.68\right)}^{2} / 9.8 - \frac{{20.68}^{2}}{2 \cdot 9.8}$

$S = 21.8 m$

This means that the stone stops momentarily at $y = 174.9 + 21.8$

$y = 196.7 m$

Now we don't have any pesky initial velocities to contend with, just a straight fall from this height:

$S = {v}_{0} t - \frac{1}{g} {t}^{2}$

${v}_{0} = 0$

As upwards is positive, falling will result in a negative displacement so

$- 196.7 = - \frac{1}{2} g {t}^{2}$

$196.7 = \frac{1}{2} g {t}^{2}$

$t = \sqrt{\frac{2 \cdot 196.7}{9.8}}$

$t = 8.45$ as required.

Aug 17, 2016

8.45s

#### Explanation:

The helicopter is asceding with a velocity $u = 20.68 \frac{m}{s}$ So the stone dropped from it will have same initial velocity as the ascending velocity of helicopter but downward gravitational force will provide it a downward acceleration (g).

Considering the point of dropping the stone from helicopter as origin we proceed as follows

If upward initial velocity be taken positive then downward acceleration (g) should be taken as negative and downard displacement (h) should also be considered negative.

$\textcolor{red}{\text{Here upward +ve and downward -ve}}$

Now calculation of time (t) of reaching ground

So we have

$u = + 20.68 \frac{m}{s}$
$g = - 9.8 \frac{m}{s} ^ 2$
$h = - 174.9 m$

t=?

Inserting these in equation of motion under gravity (comprising the variables h,u,g,t) we get

$h = u \times t + \frac{1}{2} \times g \times {t}^{2}$

$\implies - 174.9 = 20.68 \times t - \frac{1}{2} \times 9.8 \times {t}^{2.} \ldots \left(1\right)$

$\implies 4.9 {t}^{2} - 20.68 t - 174.9 = 0$

$\implies t = \frac{20.68 + \sqrt{{\left(- 20.68\right)}^{2} - 4 \cdot 4.9 \cdot \left(- 174.9\right)}}{2 \cdot 4.9}$

$\therefore t = 8.45 s$

The same equation(1) will be obtained if we reverse the direction$\textcolor{red}{\text{ i.e.upward - ive and downward + ive.}}$