Aka Posted October 16, 2006 Share Posted October 16, 2006 Hi, I need help with this question At a publicity event, Ayida, a stuntperson, will jump out of a helicopter with a jetpack on her back. The jetpack allows her to achieve a net upward acceleration of 4.4m/s^2 for a single interval of maximun lenght 10s. Ayida wants to time the use of the jetpack so that she lands with a zero velocity. a) If the helicpter is 100m hight, when should Ayida turn on her jetpack? When will she land? b) If the helicopter is 200m hight, when should Ayadia turn on her jetpack? when will she land? c) What is the maximum height from wich Ayida can jump to land with zero velocity? I don't know where to start can someone explain this problem thanks. Link to comment Share on other sites More sharing options...
apologia Posted October 17, 2006 Share Posted October 17, 2006 all i can think of is the bouncing ball problem. You know, if you bounce a ball from a building, then it goes half the distance each time... I think this is similar in the way that you're seeing how close (limit) you can come to the ground before having to go back up, then minus that distance and so on. Hope you caught any of that. Link to comment Share on other sites More sharing options...
Aka Posted October 24, 2006 Author Share Posted October 24, 2006 thanks, I'll try that Link to comment Share on other sites More sharing options...
BigMoosie Posted November 23, 2006 Share Posted November 23, 2006 Problem © appears to be the easiest, especially if you think about it in reverse. Consider the equivalent problem, a person takes of from the ground at 4.4 m per s for 10 seconds. Call this height reached A, and this final velocity B. Calculate the length of time taken to fall under gravity before velocity B is acheived, then calculate the height fallen as C. The final answer is A + C. (A) and (B) if I am not mistaken are actually incredibly tedious, I tried to calculate a function for a general case of this problem for perfect accelleration and deceleration (for a game I was making) and never came across a neat solution. Link to comment Share on other sites More sharing options...
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