1. Apr 2, 2004

### curiousguy

I have been trying to figure this out for 2 weeks.
A helicopter is begins its horizontal motion 2500 ft from the center of a targe 30 ft above the target traveling at 40mph. The target ( a pile of boxes) is stacked 15 ft high off the ground. At the appropriate time, a person is to jump out of the helicopter and hit the target. For the purpose of this problem, my professor says do not consider the horizontal speed of the jumper. I am to find the quad formula and parabola that states 1) when the person should jump to hit the target, 2) the speed of the jumper upon impact with the target. HELP! I am soooo stuck. I cannot get the quad formula for this! I did convert mph to ft per sec. Is it rational to use 16 ftpersec^2 as "g", acceleration due to gravity? When I use it, and graph the parabola it doesn't come close to hitting the target. When I try to graph the parabola from a couple of coordinates I know, it won't do it. HELP!

Last edited: Apr 2, 2004
2. Apr 2, 2004

### check

Why can't everyone just use the metric system?
Your first problem is that g is not -16ft/s^2 but more like
-32.19ft/s^2.... try that

3. Apr 2, 2004

### curiousguy

The professor has us using the falling body formula=- 1/2 32ft/s^2 + Init.Velocity + Init. Height(y intercept) That's why the -16 is there. I have narrowed down the problem I am having... its the initial velocity. Since I have been instructed to not take into account the horizontal movement of 40mph (of 58.57 ft/sec), then the parabola I come out with makes NO sense. It traces the maximum of the parabola to approx 90 ft, meaning when the guy jumps out of the helicopter, he first climbs up to 90 ft in the air, before gravity takes over and pulls him back down. Well, I know that's not right. I have spending the past 2 wks throwing stuff and jumping off of stuff just to figure out this problem!!! Oh, if only my grade weren't so important to me. But it is. Do you know what I am supposed to do about initial velocity? If I try to put in "0"x, it comes out to no solution. Arg. Thank you for the suggestion.. keep them coming.

4. Apr 2, 2004