How long in seconds was the kangaroo in the air?

  • Thread starter annyaroo
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  • #1
annyaroo
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I have a problem which is a kangaroo jumps to a vertical height of 227 cm. how long in seconds was the kangaroo in the air? so far i have 2.27m=0+1/2 (9.8)t^2 = 2.27=4.9t^2 ~ 4.9/2.27= 2.15859, and then i took the square root of 2.15859 and got 3.89 m/s. could you please tell me what i did wrong, or if i am using the right equation?
 

Answers and Replies

  • #2
zwtipp05
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The kangaroo must have an initial velocity in order to reach the height.
There are two ways to do this probelm:

You need to set up two equations:
First, the kangaroo's velocity is zero at the top of the jump so you know that the following equation is valid:
[tex]0=v_i+gt_{.5}[/tex]

Also we know that the following equation should be true:

[tex]0=v_i(2t_{.5})+\frac{1}{2}g(2t_{.5})^2[/tex]

Since the kangaroo will end at zero when it hits the ground again.
Solving the system of equations will give you t.


Alternatively, half the time of the jump will be going up and half will be going down.
So
[tex]2.27 m=\frac{1}{2}(t_{.5})^2[/tex]
 

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