Question

A helicopter starting on the ground is rising directly

Answer 1

Answer:

Oook, quick setup: pick a cartesian plane, be $(0;y(t))$the position of the helicopter as time passes, and $(x(t);0)$ your position as you start running. In my horrible sketch, the green line is the distance when you start running, the red line is the distance you need,  in general what you want is to move one end of the segment up, the other right, and stretch it.

That distance is easily found with the formula for the distance between two points in the plane, or $d=\sqrt{(\Delta x)^2+(\Delta y)^2}= \sqrt{x^2(t)+y^2(t)}$ (playing a bit with the squares). At this point we need to write down expressions for both $x(t), y(t)$ and find at what value of t we need to evaluate our distance.

For the sake of semplicity, let's start measuring times the moment you start running. Since speed is constant, both expressions will be of the form $s=vt+s_0$

The equation for the runner becomes simply $x(t)=10t$, since you haven't moved until the heli is high enough.

The equation of the helicopter is $y=25t+30$ since the heli is 30ft above ground when you start running.

Finally, we need to know how many seconds have passed when the heli is at 60 ft above ground. that happens when $60=25t+30 \rightarrow t=\frac{30}{25} = 1.20s$. in this time, you are at  $x(1.2)=10(1.2)=12ft$ from the origin.

Plugging it in the distance formula you get: $\sqrt{12^2+60^2}= \sqrt{144+3600}=4\sqrt {234}\approx 61.2ft$