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Thai Pham

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About Thai Pham

  • Birthday 01/16/1993


  • Biography
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    Milpitas, California, United States
  • Interests
    Running Without Shoes, Sleep Without Pillow, Being Awesome, Stay Up Late Ho Chi Minh, Rest In Peace,

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  1. 3.90 ... CP A rocket designed to place small payloads into orbit is carried to an altitude of 12.0 km above sea level by a converted airliner. When the airliner is flying in a straight line at a constant speed of 850km/h, the rocket is dropped. After the drop, the airliner maintains the same altitude and speed and continues to fly in a straight line. The rocket falls for a brief time, after which its rocket motor turns on. Once its rocket motor is on, the combined effects of thrust and gravity give the rocket a constant acceleration of magnitude 3.00g directed at an angle of 30 degree above the horizontal. For reasons of safety, the rocket should be at least 1.00 km in front of the airliner when it climbs through the airliner’s altitude. Your job is to determine the minimum time that the rocket must fall before its engine starts. You can ignore air resistance. Your answer should include (i) a diagram showing the flight paths of both the rocket and the airliner, labeled at several points with vectors for their velocities and accelerations; (ii) an x-t graph showing the motions of both the rocket and the airliner; and (iii) a y-t graph showing the motions of both the rocket and the airliner. In the diagram and the graphs, indicate when the rocket is dropped, when the rocket motor turns on, and when the rocket climbs through the altitude of the airliner I have no idea where to start? Here is the diagram that draw, not sure if it is accurate
  2. This is a very good question and I got the same answer as your teacher :eagerness: The first thing I should do is drawing the diagram to see the actual picture of the problem Next, since this is Projectile Motion problem, this problem require 2 components, x and y Now, no the given, so you can choose which equation to choose. For this, we choose the following equation [TABLE="class: maths"] [TR] [TD="align: right"]x = [/TD] [TD="align: left"]x0 + v0Δt + ½aΔt2 [/TD] [/TR] [/TABLE] Now, set the two components for x and y y-component: Yf - Yi = V(iy) t + 12 ayt2 You know the origin is where the ball is hit which makes yf=0 and yi which stand for y final equal to -2.4m. v initial is zero because the ball is throwing side away so imagine like dropping the ball. The acceleration is 9.8. Now sub it in, you get the amount of time that it takes to hit the ground. x-component: v initial is 30m/s acceleration is 0 since the direction is going side way just plug it in to the x component. After you have the equation set up, plug in the t which give you the answer for B. The distance that it hits the ground For part A just use 12m away, so you want to know when the ball flies pass at 12m. Simple, use the x component equation. make x-xinitial=12m. Now you have the time which is 0.4second The last step is plug into y component to find how high was the ball almost reach to the ground which is 0.784m Now you can get the answer by simply take 2.4m-0.784m=20.7m I hope this help, if you have trouble doing this, should check out more videos tutorials :star:
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