A helicopter is ascending vertically with a speed of 2.50 m/s. At a height of 145 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground?

Guest Jan 9, 2016

#11**+20 **

**A helicopter is ascending vertically with a speed of 2.50 m/s. At a height of 145 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground?**

**\(\begin{array}{rcl} h &=& h_0+v_0\cdot t-\frac12\cdot g\cdot t^2 \qquad | \qquad h = 0\ m \\ 0 &=& h_0+v_0\cdot t-\frac12\cdot g\cdot t^2 \\ -\frac12\cdot g\cdot t^2 + v_0\cdot t +h_0 &=& 0\\\\ \boxed{~ \begin{array}{rrcl} ax^2+bx+c &=& 0 \\ x &=& {-b \pm \sqrt{b^2-4ac} \over 2a} \end{array} ~}\\\\ \begin{array}{rcl} a=-\frac12\cdot g\qquad b=v_0 \qquad c = h_0 \\ \end{array}\\ t_{1,2} &=& {-v_0 \pm \sqrt{v_0^2-4(-\frac12\cdot g)\cdot h_0} \over 2(-\frac12\cdot g)} \\ t_{1,2} &=& {-v_0 \pm \sqrt{v_0^2+2g\cdot h_0} \over -g } \\ t_{1,2} &=& {v_0 \pm \sqrt{v_0^2+2g\cdot h_0} \over g } \\\\ \begin{array}{rcl} v_0= 2.50\ \frac{m}{s} \qquad h_0 = 145\ m \qquad g= 9.81\ \frac{m}{s^2}\\ \end{array}\\ t_{1,2} &=& {2.50 \pm \sqrt{2.50^2+2\cdot 9.81\cdot 145} \over 9.81 } \\\\ t_{1,2} &=& \frac{2.50 \pm 149.903802487} { 9.81 } \\ t_1 &=& \frac{2.50 + 53.3961609107} { 9.81 } \\ t_1 &=& 5.69787572994\ s\\\\ t_2 &=& \frac{2.50 - 53.3961609107} { 9.81 } \\ t_2 &=& -5.18819173402\ s \quad \text{ negative time, no solution} \end{array}\)**

**t = 5,7 s**

heureka
Jan 11, 2016

#1**+10 **

A helicopter is ascending vertically with a speed of 2.50 m/s. At a height of 145 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground?

Use Galileo's famous formula for falling bodies: D=1/2.g.t^2, where D=distance the object falls, g=9.81 m/sec^2, the pull of gravity, t=time it takes the object to hit the ground.

Guest Jan 9, 2016

#2**+5 **

I believe that the helicopter's rate of ascent is irrelevant......at the time of release, the package is 145m above the ground instantaneously....so we have

145 = 4.9t^2 divide both sides by 4.9

145/4.9 = t^2 take the square root of both sides

t = about 5.44 s

CPhill
Jan 9, 2016

#3**+5 **

Actually CPhill it do make a difference. The package is moving up at 2.5 meters per second when it leave the helicopter. So it continue to move as gravity slows it down then it stop and goes down so it is higher up when it starts down. I not figure out how to do that yet. But I working on it :)

Dragonlance
Jan 9, 2016

#4**+5 **

You could be correct, Dragonlance......if so....I think we might have this formula :

0 = -4.9t^2 + 2.5t + 145

Solving this for t produces.....

t = 5.7 s

I'd like to have some "Physics" person look at this.....I'm not sure which is correct [ if either one is !!! ]

CPhill
Jan 9, 2016

#5**+5 **

OK CPhill this is the equation I get. It is a quadratic so it will have 2 answers. Right?

0 m = 0.5(-9.8 m/s^2)t^2 + (2.50 m/s)t + 145m

----

Our equations are the same so this is probably right.

Dragonlance
Jan 9, 2016

#6**+5 **

Yeah...Dragonlance.......it's the same equation as the second one I used......since t is positve, I beileve that 5.7 s will be the correct answer....

Here's a graph of the solution.....https://www.desmos.com/calculator/uojhmuqlau

[ If it* IS* the solution.....I'm still not sure about this one....!!!]

CPhill
Jan 9, 2016

#7**+5 **

This is interesting. I was looking at your graph and I sure the answer is right but the graph do not show what really happen. The graph show a slowdown of the gravity it do not show the package going higher.

How would do you write the equation to show the package going higher then coming down?

Dragonlance
Jan 9, 2016

#8**+5 **

Actually, Dragonlance, the graph * does* show that :

At t = 0, the package is at 145 m

However......at t = about.2551 seconds, the package is at about 145.319 m.......and this is the vertex of the parabola = the max height

Look at the same "blown-up" graph with the points added : https://www.desmos.com/calculator/zjuovdmo2h

CPhill
Jan 9, 2016

#9**+5 **

Thank you CPhill I do see it now. This really cool to see these graphs show what happen in the real world.

Next come the air resistance. All the (hot) air will make the equation twice as long and 4 times as hard to figure out. hahaha

Dragonlance
Jan 9, 2016

#10**+5 **

I think I already generated enough hot air over this problem.....let's just let it go and call it all good.......LOL!!!!

CPhill
Jan 9, 2016

#11**+20 **

Best Answer

**\(\begin{array}{rcl} h &=& h_0+v_0\cdot t-\frac12\cdot g\cdot t^2 \qquad | \qquad h = 0\ m \\ 0 &=& h_0+v_0\cdot t-\frac12\cdot g\cdot t^2 \\ -\frac12\cdot g\cdot t^2 + v_0\cdot t +h_0 &=& 0\\\\ \boxed{~ \begin{array}{rrcl} ax^2+bx+c &=& 0 \\ x &=& {-b \pm \sqrt{b^2-4ac} \over 2a} \end{array} ~}\\\\ \begin{array}{rcl} a=-\frac12\cdot g\qquad b=v_0 \qquad c = h_0 \\ \end{array}\\ t_{1,2} &=& {-v_0 \pm \sqrt{v_0^2-4(-\frac12\cdot g)\cdot h_0} \over 2(-\frac12\cdot g)} \\ t_{1,2} &=& {-v_0 \pm \sqrt{v_0^2+2g\cdot h_0} \over -g } \\ t_{1,2} &=& {v_0 \pm \sqrt{v_0^2+2g\cdot h_0} \over g } \\\\ \begin{array}{rcl} v_0= 2.50\ \frac{m}{s} \qquad h_0 = 145\ m \qquad g= 9.81\ \frac{m}{s^2}\\ \end{array}\\ t_{1,2} &=& {2.50 \pm \sqrt{2.50^2+2\cdot 9.81\cdot 145} \over 9.81 } \\\\ t_{1,2} &=& \frac{2.50 \pm 149.903802487} { 9.81 } \\ t_1 &=& \frac{2.50 + 53.3961609107} { 9.81 } \\ t_1 &=& 5.69787572994\ s\\\\ t_2 &=& \frac{2.50 - 53.3961609107} { 9.81 } \\ t_2 &=& -5.18819173402\ s \quad \text{ negative time, no solution} \end{array}\)**

**t = 5,7 s**

heureka
Jan 11, 2016