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?
+26367 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
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.
+128408 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
+1315 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 :)
+128408 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 !!! ]
+1315 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.
+128408 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....!!!]
+1315 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?
+128408 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
+1315 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
+128408 I think I already generated enough hot air over this problem.....let's just let it go and call it all good.......LOL!!!!
+26367 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
