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Freely Falling Objects

  1. Sep 18, 2005 #1
    Hey Everybody,

    I am a little perplexed on how to answer this problem, I got an answer, but it doesn't seem to fit within the context of the problem. I don't know if it is because I am using the wrong equation, or if I am just plugging the numbers in wrong:

    A rocket is fired vertically upward with an initial velocity of 80.0 m/s. It accelerates upward at 4.00 m/s^2 until it reaches an altitude of 1000 m. At that point, its engines fail and the rocket goes into free flight, with an acceleration of -9.8 m/s^2. (a) How long is the rocket in motion? (b) What is its maximum altitude? (c) What is its velocity just before it collides with the earth?

    Also, I was wondering if someone would mind checking this problem for me:

    The Height of a helicopter above the ground is given by h=3.00t^3, (h is in m., t is in s.) After 2.00 s, the helicopter releases a small mailbag. How long after its release does the mailbag reach the ground?


    Thanks for the help!!
  2. jcsd
  3. Sep 18, 2005 #2
    i)use v^2=u^2+2as to get the velocity when its at 1000m.
    ii)Then use s=ut+1/2*at^2 to get the time of flight. The "u" in the formula is the velocity at the altidue 1000m which you determined in (i). s=-1000 and a=-9.8. find t. this will be the time of flight after it switched off the engines. to get the time b4 it did this use the same forumla only sub in s=1000 a=4 and u=0. add the two times together

    iii) for max altidude use same equation as in (i) onlys sub in v=0 a=-9.8 (as opposed to 4) and u= what ever you got in (i) originally. find s and add 100 to it

    iv)for final velocity use v=u+at where u is the velocity after the engines are switch off (determine in (i)) a=-9.8 and t is the time off flight after engines are switched off.
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