A helicopter poliet drops a package from 200ft above the ground and rising at a speed of 20ft/sec. how long will it take for the package to hit the ground?

Question

kmeis002 · Answer

A couple ways to approach this, but we can do it all algebraically from the kinematic equations.

$$s(t) = s(t_0)+v(t_0)t+\frac{1}{2}a(t_0)t^2 $$ where s= position, v = velocity, and a = acceleration and t_0 is the initial time. For this problem, we can arbitrarily choose t_0 = 0, for simplicity, and a assume to be a constant of -9.81 (earth's gravitational acceleration.) Also, the package STARTED at 200 so s(0) = 200 and s(t) = 0 (hits the ground). We are also given an initial speed v(0) = 20. Plug it in:

$$ 0 = 200 + 20t -\frac{1}{2}9.81t^2$$

Now we just solve this quadratic for t.

anonymous · Answer

would i take the derivative of that?

anonymous · Answer

@kmeis002

kmeis002 · Answer

Ah, for this you just need to use the quadratic equation or factoring, t will have t values, one will make sense with respect to the problem.

anonymous · Answer

hmm okay let me try it out that way and see what i get

anonymous · Answer

um im having problems with simplifying it .-.

kmeis002 · Answer

we can identify a, b c as
$$a = -\frac{9.81}{2} \ b = 20 \ c = 200 \ \ 
t = \frac{-b\pm\sqrt{b^2-4ac}}{2a} = \frac{-20\pm\sqrt{(20)^2-4(200)\left ( \frac{-9.81}{2} \right )}}{2\left ( \frac{-9.81}{2} \right )}  $$

anonymous · Answer

okay yeah i did that but i got a nonreal answer

kmeis002 · Answer

Check your signs, -4ac and c is negative so the sqrt should be strictly positive

kmeis002 · Answer

*a is negative

anonymous · Answer

okay i got 4320 under the sqrt

kmeis002 · Answer

Good, so when you compute the sqrt and the final two answers, one t should be negative and one will be positive. Only one makes sense in the context of this problem