Im having trouble with a few preoblems.

  • Thread starter thschica
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  • #1
thschica
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Im having trouble with a few problems.

1.A kangaroo jumps to a vertical height of 2.5 m. How long was it in the air before returning to Earth?
? s ( On this problem I keep getting a negative number.I know that this on is wrong for sure. What equation should I be using?)

3. A helicopter is ascending vertically with a speed of 6.00 m/s. At a height of 170 m above the Earth, a package is dropped from a window. How much time does it take for the package to reach the ground?
?s ( I have tried three equations but I still can figure out exactly which one I should be using)

10. A stone is thrown vertically upward with a speed of 13.0 m/s from the edge of a cliff 72.0 m high
(a) How much later does it reach the bottom of the cliff?
? s
(b) What is its speed just before hitting?
?m/s
(c) What total distance did it travel?
?m
(Once again I am getting negatives!)

25. A rock is dropped from a sea cliff and the sound of it striking the ocean is heard 6.6 s later. If the speed of sound is 340 m/s, how high is the cliff?
? m (This one I am not sure what I am doing wrong)
 
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Answers and Replies

  • #2
HallsofIvy
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Did you check the "rules" for this Forum (see "Read this before posting"). We aren't going to do your homework for you. Show us what you have tried or at least what ideas you might have about the problem.
 
  • #3
Curious3141
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This link should give you all the info you need to start solving the problems.

You have to show your work before asking for help, that's the rule here. :)
 
  • #4
thschica
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I know how to do #10 but #1 is the one I am having the most trouble on.Can someone help me?
 
  • #5
Doc Al
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thschica said:
1.A kangaroo jumps to a vertical height of 2.5 m. How long was it in the air before returning to Earth?
? s ( On this problem I keep getting a negative number.I know that this on is wrong for sure. What equation should I be using?)
Show what you did to get a negative number.
 
  • #6
thschica
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I said 2.5m(t)+.5(-4.80)t^2
 
  • #7
TD
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Why did you use -4.80?
 
  • #8
Doc Al
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thschica said:
I said 2.5m(t)+.5(-4.80)t^2
Were you trying to use this formula?
[itex]x = x_0 + v_0 t + .5 a t^2[/itex]
 
  • #9
thschica
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yes I was.I meant 9.8
 
  • #10
Doc Al
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OK, so how were you planning to use that formula? If you start at ground level, you'll need to figure out the initial speed. But if you realize that the kangaroo's motion going up is a mirror image of its motion going down, you can apply the formula starting at the top of the motion where the speed is zero. Give it another shot. (Define all your values before "plugging in".)
 

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