1. The problem statement, all variables and given/known data The height of a helicopter above the ground is given by h is in meter and t is in seconds. After 2.00 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground? 2. Relevant equations v_{xf}= v_{xi}+a_{x}*t v_{x,avg}= (v_{xi} +v_{xf})/2 x_{f}=x_{i}+1/2(v_{xi} +v_{xf})*t x_{f}=x_{i}+v_{xi}*t+1/2a*t^{2} v_{xf}^{2}= v_{xi}^{2}+2*a(x_{f}-x_{i}) 3. The attempt at a solution What I did was put 2 sec into the height equation, then use the above bolded equation, where i put the constant of gravit as the acceleration and v_{xi} as 0 and x_{i} as h(2) and x_{f} as 0, but the webassign online says i got it wrong and i did it like 5 times. Can someone tell me their thought process and the answer? Thanks!
Why put in 2? Is there something you left out of the problem statement? The time is given directly by h = 1/2*g*t^{2}
Sorry, the equation given was h=3.00t^{3} I put in 2 seconds to find the height at which the helicopter releases the mailbag, and then put in that height as x_{i}.
h = 3* t^{3}? Whoa. That's a whole other ocean. It's got a vertical velocity component up as well as being in free fall. What do you figure is the vertical velocity when it is released?
From that equation it would. dy/dt (vertical velocity) is certainly positive. Does the problem say anything other than that it is just released. Are taking derivatives something you are supposed to be able to do in this course? Do you know how to take dy/dt of y = 3 t^{3} ?
No, everything that I stated in the first post is all the info I have. The course im taking is a calculus based class, and I have already taken calc 3 and diff EQ, so yea i know how to take derivatives. But in class all we have been really using so far are just the formulas that I gave in the first post. If I did use dh/dt, where dh/dt= 9t^{2}, then I guess I could use one of the forumlas where I could use v_{f}, but i still am confused.
OK then great. That's what your vertical velocity is. dh/dt = V_{y} = 9t^{2} and at 2 seconds that value is your upward velocity.
Figure then the height that it will continue to go up. V^{2}/(2*a) = Y That gives you it's maximum height (when you add the height it was dropped at). But you need time. So figure from the first calculated height how long it took. Y = 1/2 a* t^{2} That's time to max height after release. Now take the total height and put it in the same equation again and you have the time to fall. Time up + time down = total time. VoilĂ .
But what does that have to do with dropping the bag? If you said that at 2 sec thats my value of the upward velocity, so then that is v_{i}. So to find where the bag reaches its maximum height, v= 0. So the upward velocity at 2 sec is 36 m/s, so to find the time where the bag reaches the maximum height, you have to use v_{yf}=v_{yi}+a_{y}t. When i do that I get t =3.67 sec. But then how do I find the maximum height? Am i in the right direction?